Chapter 14

Functions of a Complex Variable

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Section 14.1The Cauchy--Riemann Equations[16/75%]
Problem 5371

Find the analytic function w(z)w(z) from the following conditions: u=x2−y2+xyu = x^{2}-y^{2}+xy; w(0)=0w(0) = 0.

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Problem 5372

Find the analytic function w(z)w(z) from the following conditions: u=x3+6x2y−3xy2−2y3u = x^{3}+6x^{2}y-3xy^{2}-2y^{3}; w(0)=0w(0) = 0.

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Problem 5373

Find the analytic function w(z)w(z) from the following conditions: v=yx2+y2v = \frac{y}{x^{2}+y^{2}}; w(2)=0w(2) = 0.

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Problem 5374

Find the analytic function w(z)w(z) from the following conditions: u=ex(xcos⁡y−ysin⁡y)u = e^{x}(x\cos y - y\sin y); w(0)=0w(0) = 0.

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Problem 5375

Find the analytic function w(z)w(z) from the following conditions: u=2sin⁡2xe2y+e−2y−2cos⁡2xu = \frac{2\sin 2x}{e^{2y}+e^{-2y}-2\cos 2x}.

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Problem 5376

Find the analytic function w(z)w(z) from the following conditions: u=φ(yx)u = \varphi \left(\frac{y}{x}\right), where φ\varphi is some function.

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Problem 5377

Find the analytic function w(z)w(z) from the following conditions: u=F(x+x2+y2)u = F(x+\sqrt{x^{2}+y^{2}}).

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Problem 5378

Find the analytic function w(z)w(z) from the following conditions: v=f(x2+y2)v = f(x^{2}+y^{2}).

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Problem 5379

Prove that the equation u(x,y)=Cu(x,y) = C can represent a family of lines of equal potential only when the ratio of the quantities

∂2u∂x2+∂2u∂y2and(∂u∂x)2+(∂u∂y)2 \frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} \quad \text{and} \quad \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}

is a function of uu alone.

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Problem 5380

Prove that the curves xn+yn=Cx^{n}+y^{n} = C can be lines of equal potential only for n=1n = 1 and for n=2n = 2.

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Problem 5381

Find the analytic function w(z)w(z) from the conditions: the lines Re(w(z))=c\mathrm{Re}\left(w(z)\right) = c are the circles x2+y2=a2x^{2}+y^{2} = a^{2}.

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Problem 5382

Find the analytic function w(z)w(z) from the conditions: the lines ∣w(z)∣=c\left|w(z)\right| = c are the circles x2+y2=axx^{2}+y^{2} = ax.

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Problem 5383

Find the analytic function w(z)w(z) from the conditions: the lines ∣w(z)∣=c\left|w(z)\right| = c are the lemniscates r2=a2sin⁡2φr^{2} = a^{2}\sin 2\varphi.

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Problem 5384

Find the analytic function w(z)w(z) from the conditions: the lines arg⁡w(z)=c\arg w(z) = c are the circles x2+y2=a2x^{2}+y^{2} = a^{2}.

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Problem 5385

Prove the identity:

∂2∣w(z)∣∂x2+∂2∣w(z)∣∂y2=∣w′(z)∣2∣w(z)∣. \frac{\partial ^{2} \left|w(z)\right|}{\partial x^{2}}+\frac{\partial ^{2} \left|w(z)\right|}{\partial y^{2}} = \frac{\left|w'(z)\right|^{2}}{\left|w(z)\right|}.
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Problem 5386

The function w=f(z)w = f(z) is regular in a simply connected domain bounded by the curve ∣w(z)∣=c\left|w(z)\right| = c. Show that arg⁡w\arg w varies monotonically as zz traverses the curve ∣w(z)∣=c\left|w(z)\right| = c in the positive direction.

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Section 14.2Singular Points of a Function. Taylor and Maclaurin Series[36/11%]
Problem 5387

Find the singular points of the function zz2+1\dfrac {z}{z^{2}+1}.

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Problem 5388

Find the singular points of the function 1z−z3\dfrac {1}{z-z^{3}}.

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Problem 5389

Find the singular points of the function 1sin⁡z\dfrac {1}{\sin z}.

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Problem 5390

Find the singular points of the function 1ez−1−1z\dfrac {1}{e^{z}-1}-\dfrac {1}{z}.

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Problem 5391

Find the singular points of the function sin⁡1z\sin \dfrac {1}{z}.

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Problem 5392

Find the singular points of the function 1−z2\sqrt{1-z^{2}}.

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Problem 5393

Find the singular points of the function 1−z3\sqrt{1-z^{3}}.

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Problem 5394

Construct the lines of equal modulus and equal argument for the function w=e1zw = e^{\frac{1}{z}}.

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Problem 5395

Show that for any A≠0A \neq 0 the equation e1/z=Ae^{1/z} = A has infinitely many roots in any given neighborhood of the point z=0z = 0.

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Problem 5396

Show that in the same region the equation cos⁡1z=A\cos \dfrac {1}{z} = A has infinitely many roots for any value of AA.

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Problem 5397

Study the change in arg⁡z−an\arg \sqrt[n]{z-a} as the point zz traces a simple closed contour going around the point z=az = a in the positive direction.

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Problem 5398

Study the change in the function 1−z2\sqrt{1-z^{2}} as the point zz traces a simple closed contour of one of three kinds:

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(a)

going around one of the points z=±1z = \pm 1,

(b)

going around both of these points,

(c)

leaving these points outside the contour.

Problem 5399

Find the value at the point z=−ez = -e of the function w=zz=ezln⁡zw = z^{z} = e^{z\ln z}, if ln⁡1=0\ln 1 = 0 and if ww varies continuously as zz moves in the upper half-plane, or as zz moves in the lower half-plane.

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Problem 5400

Find the value at z=e−φiz = e^{-\varphi i} of the branch of the function z−eφi\sqrt{z-e^{\varphi i}} which at z=0z = 0 has the value ieφ2iie^{\frac{\varphi }{2}i} and which is continued continuously as zz moves along the radius from the origin to the point z=e−φiz = e^{-\varphi i}.

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Problem 5401

In the zz-plane a cut is made along the lines: y=0y = 0; ∣x∣>1\left|x\right| > 1 and x2+y2=1x^{2}+y^{2} = 1; y>0y > 0; the function ln⁡(1−z2)\ln (1-z^{2}) is zero at z=0z = 0. Find its value at z=3z = 3.

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Problem 5402

Show that the function w=ln⁡(z+z2−1)w = \ln (z+\sqrt{z^{2}-1}) is single-valued in the plane with cuts along the real axis for −∞<x<−1-\infty < x < -1 and 1<x<∞1 < x < \infty. Find the values of ww at z=+iz = +i and z=−iz = -i, given that on the upper edge of the right-hand cut ww is positive.

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Problem 5403

Using the series expansion for 1cos⁡z\dfrac {1}{\cos z} and zctg⁡zz\ctg z, prove the identity:

ln⁡tg⁡(π4+z2)=∑n=0∞En(2n+1)!z2n+1;∣z∣<π2; \ln \tg \left(\frac{\pi }{4}+\frac{z}{2}\right) = \sum _{n=0}^{\infty } \frac{E_{n}}{(2n+1)!}z^{2n+1}; \quad \left|z\right| < \frac{\pi }{2};

EnE_{n} are the Euler numbers (E0=1E_{0} = 1).

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Problem 5404

Using the series expansion for 1cos⁡z\dfrac {1}{\cos z} and zctg⁡zz\ctg z, prove the identity:

tg⁡z=∑n=1∞22n(22n−1)Bn(2n)!z2n−1;∣z∣<π2; \tg z = \sum _{n=1}^{\infty } \frac{2^{2n}(2^{2n}-1)B_{n}}{(2n)!}z^{2n-1}; \quad \left|z\right| < \frac{\pi }{2};

BnB_{n} are the Bernoulli numbers.

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Problem 5405

Using the series expansion for 1cos⁡z\dfrac {1}{\cos z} and zctg⁡zz\ctg z, prove the identity:

zsin⁡z=1+∑n=1∞(22n−2)Bn(2n)!z2n;∣z∣<π. \frac{z}{\sin z} = 1+\sum _{n=1}^{\infty } \frac{(2^{2n}-2)B_{n}}{(2n)!}z^{2n}; \quad \left|z\right| < \pi .
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Problem 5406

Using the series expansion for 1cos⁡z\dfrac {1}{\cos z} and zctg⁡zz\ctg z, prove the identity:

ln⁡sin⁡zz=−∑n=1∞22n−1Bnn(2n)!z2n;∣z∣<π. \ln \frac{\sin z}{z} = -\sum _{n=1}^{\infty } \frac{2^{2n-1}B_{n}}{n(2n)!}z^{2n}; \quad \left|z\right| < \pi .
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Problem 5407

Using the series expansion for 1cos⁡z\dfrac {1}{\cos z} and zctg⁡zz\ctg z, prove the identity:

ln⁡cos⁡z=−∑n=1∞22n−1(22n−1)Bnn⋅(2n)!z2n;∣z∣<π2. \ln \cos z = -\sum _{n=1}^{\infty } \frac{2^{2n-1}(2^{2n}-1)B_{n}}{n\cdot (2n)!}z^{2n}; \quad \left|z\right| < \frac{\pi }{2}.
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Problem 5408

Find the first six coefficients in the expansion of the functions, and determine the domains of convergence of these series:

w1=zln⁡(1+z);w2=zarctg⁡z. w_{1} = \frac{z}{\ln (1+z)}; \quad w_{2} = \frac{z}{\arctg z}.
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Problem 5409

Expand in a series of powers of zz the branch of the function z+i\sqrt{z+i} whose value at z=0z = 0 is 1+i2\dfrac {1+i}{\sqrt{2}}.

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Problem 5410

Find the circle of convergence for the expansion in powers of zz of the functions sin⁡πz2sin⁡πz\dfrac {\sin \pi z^{2}}{\sin \pi z} and cos⁡2πz2cos⁡πz\dfrac {\cos 2\pi z^{2}}{\cos \pi z}.

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Problem 5411

Expand the function z(z2−4)(z2−1)\dfrac {z}{(z^{2}-4)(z^{2}-1)} in a Laurent series for 1<∣z∣<21 < \left|z\right| < 2.

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Problem 5412

Expand the function 1(z2−1)(z2−4)2\dfrac {1}{(z^{2}-1)(z^{2}-4)^{2}} in a Laurent series for 1<∣z∣<21 < \left|z\right| < 2.

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Problem 5413

Expand the function 1(z2−1)(z2−4)2\dfrac {1}{(z^{2}-1)(z^{2}-4)^{2}} in a Laurent series for ∣z∣>2\left|z\right| > 2 (the same function as in the previous problem, for a different domain condition).

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Problem 5414

Expand the function ctg⁡z\ctg z in a Laurent series:

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(a)

for ∣z∣<π\left|z\right| < \pi;

(b)

for π<∣z∣<2π\pi < \left|z\right| < 2\pi.

Problem 5415

Expand the function 1z−2ln⁡z−iz+i\dfrac {1}{z-2}\ln \dfrac {z-i}{z+i} in a Laurent series for 1<∣z∣<21 < \left|z\right| < 2, taking the value of the logarithm at z=2iz = 2i to be real.

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Problem 5416

Expand the function 1z−2ln⁡z−iz+i\dfrac {1}{z-2}\ln \dfrac {z-i}{z+i} in a Laurent series for ∣z∣>2\left|z\right| > 2 (the same function as in the previous problem, for a different domain condition, taking the value of the logarithm at z=2iz = 2i to be real).

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Problem 5417

Expand the function z5e1/zz^{5}e^{1/z} in a Laurent series.

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Problem 5418

Expand the function ez+1/ze^{z+1/z} in a Laurent series.

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Problem 5419

Prove the theorem: if the series ∑n=0∞anzn\sum_{n=0}^{\infty } a_{n}z^{n} has only one singular point on the boundary of its circle of convergence, namely a pole of order mm, then the identity an=Anm−1z0n(1+ε(n))a_{n} = \dfrac {An^{m-1}}{z_{0}^{n}}\left(1+\varepsilon (n)\right) holds, where ε(n)→0\varepsilon (n) \to 0 as n→∞n \to \infty, AA is a constant, and z0z_{0} is the pole.

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Problem 5420

Prove the theorem: if the radius of convergence of the power series ∑anzn\sum a_{n}z^{n} equals one, and the only singular points on the circle of convergence are poles whose order does not exceed mm, then the coefficients of the series do not exceed Anm−1An^{m-1}, where AA is a constant.

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Problem 5421

Prove the theorem: if for the series ∑anzn\sum a_{n}z^{n} all an⩾0a_{n} \geqslant 0, and the radius of the disk equals 11, then z=1z = 1 is a singular point.

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Remark.
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Proof by contradiction, based on the expansion of the sum of the series in powers of z−az-a, where 0<a<10 < a < 1.

Problem 5422

Prove the theorem: if the radius of convergence of the series ∑anzn\sum a_{n}z^{n} and ∑Re(an)⋅zn\sum \mathrm{Re}\left(a_{n}\right)\cdot z^{n} equals 11 and all Re(an)⩾0\mathrm{Re}\left(a_{n}\right) \geqslant 0, then the point z=1z = 1 is a singular point for the sums of these series.

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Section 14.3Residues and Their Applications[66/42%]
Problem 5423

Find the residues of the function 1z3−z5\dfrac {1}{z^{3}-z^{5}} at the points z=0z = 0, z=±1z = \pm 1.

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Problem 5424

Find the residues of the function z2(z2+1)2\dfrac {z^{2}}{(z^{2}+1)^{2}} at the points z=±iz = \pm i.

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Problem 5425

Find the residues of the function z2n(z+1)n\dfrac {z^{2n}}{(z+1)^{n}} at the points z=−1z = -1 and z=∞z = \infty.

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Problem 5426

Find the residue of the function ezln⁡z−αz−βe^{z}\ln \dfrac {z-\alpha }{z-\beta } at the point z=∞z = \infty.

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Problem 5427

Compute the integral ∫dzz4+1\displaystyle \int \frac{dz}{z^{4}+1} along the circle x2+y2=2xx^{2}+y^{2} = 2x, taking the path counterclockwise.

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Problem 5428

Compute the integral ∫dz(z−1)2(z2+1)\displaystyle \int \frac{dz}{(z-1)^{2}(z^{2}+1)} along the circle x2+y2=2x+2yx^{2}+y^{2} = 2x+2y, taking the path counterclockwise.

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Problem 5429

Compute the integral ∫dz(z2−1)2(z−3)2\displaystyle \int \frac{dz}{(z^{2}-1)^{2}(z-3)^{2}} along the astroid x23+y23=223x^{\frac{2}{3}}+y^{\frac{2}{3}} = 2^{\frac{2}{3}}, taking the path counterclockwise.

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Problem 5430

Compute the integral ∫∣z∣=1dz(z−a)n(z−b)n\displaystyle \int_{\left|z\right|=1} \frac{dz}{(z-a)^{n}(z-b)^{n}}; ∣a∣<1\left|a\right| < 1, ∣b∣<1\left|b\right| < 1, taking the path counterclockwise.

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Problem 5431

Compute the integral ∫dz(z2−1)(z3−1)\displaystyle \int \frac{dz}{(z^{2}-1)(z^{3}-1)} along the circle ∣z∣=r≠1\left|z\right| = r \neq 1, taking the path counterclockwise.

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Problem 5432

Compute the integral ∫dzz3(z10−2)\displaystyle \int \frac{dz}{z^{3}(z^{10}-2)} along the circle ∣z∣=2\left|z\right| = 2, taking the path counterclockwise.

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Problem 5433

Compute the integral ∫z2e2πiz3−1 dz\displaystyle \int \frac{z^{2}}{e^{2\pi i z^{3}}-1}\, dz along the edge of the half-circle ∣z∣=R\left|z\right| = R, under the conditions n<R3<n+1n < R^{3} < n+1, where nn is an integer, taking the path counterclockwise.

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Problem 5434

Find the integral ∫dt(z4+1)z2+1\displaystyle \int \frac{dt}{(z^{4}+1)\sqrt{z^{2}+1}}, taken along the parabola x=y2x = y^{2} in the direction of increasing yy, taking the value of z2+1\sqrt{z^{2}+1} to equal one at z=0z = 0.

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Problem 5435

For what relation between aa and bb will the function

∫0tatsin⁡t+bcos⁡tt2 dt \int _{0}^{t} \frac{at\sin t+b\cos t}{t^{2}}\, dt

be single-valued?

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Problem 5436

Compute by contour integration the integral ∫−∞∞x2+1x4+1 dx\displaystyle \int_{-\infty }^{\infty } \frac{x^{2}+1}{x^{4}+1}\, dx.

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Problem 5437

Compute by contour integration the integral ∫−∞∞x4+1x6+1 dx\displaystyle \int_{-\infty }^{\infty } \frac{x^{4}+1}{x^{6}+1}\, dx.

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Problem 5438

Compute by contour integration the integral ∫−∞∞x2(x2+a2)2 dx\displaystyle \int_{-\infty }^{\infty } \frac{x^{2}}{(x^{2}+a^{2})^{2}}\, dx.

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Problem 5439

Compute by contour integration the integral ∫−∞∞dx(x2+a2)2(x2+b2)\displaystyle \int_{-\infty }^{\infty } \frac{dx}{(x^{2}+a^{2})^{2}(x^{2}+b^{2})}.

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Problem 5440

Compute by contour integration the integral ∫−∞∞x2m1+x2n dx\displaystyle \int_{-\infty }^{\infty } \frac{x^{2m}}{1+x^{2n}}\, dx; m<nm < n.

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Problem 5441

Compute by contour integration the integral ∫02πdφcos⁡φ+a\displaystyle \int_{0}^{2\pi } \frac{d\varphi }{\cos \varphi +a}; a>1a > 1 (set z=eiφz = e^{i\varphi }).

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Problem 5442

Compute by contour integration the integral ∫02πcos⁡22φ1−2pcos⁡φ+p2 dφ\displaystyle \int_{0}^{2\pi } \frac{\cos^{2} 2\varphi }{1-2p\cos \varphi +p^{2}}\, d\varphi; ∣p∣<1\left|p\right| < 1.

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Problem 5443

Compute by contour integration the integral ∫02πcos⁡23φ1−2pcos⁡2φ+p2 dφ\displaystyle \int_{0}^{2\pi } \frac{\cos^{2} 3\varphi }{1-2p\cos 2\varphi +p^{2}}\, d\varphi; ∣p∣<1\left|p\right| < 1.

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Problem 5444

Compute by contour integration the integral ∫0πctg⁡(x−a) dx\displaystyle \int_{0}^{\pi } \ctg (x-a)\, dx; Im(a)≠0\mathrm{Im}\left(a\right) \neq 0.

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Problem 5445

Prove the identity lim⁡r→∞∫CemizQ(z) dz=0\displaystyle \lim_{r\to \infty } \int_{C} e^{miz}Q(z)\, dz = 0, where CC is the half-circle z=reφiz = re^{\varphi i}; 0⩽φ⩽π0 \leqslant \varphi \leqslant \pi; m>0m > 0, and the function Q(z)Q(z) tends uniformly to zero as ∣z∣→∞\left|z\right| \to \infty in the sector 0⩽φ⩽π0 \leqslant \varphi \leqslant \pi and is regular in the upper half-plane. (Jordan's Lemma.)

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Problem 5446

Compute the integrals I1=∫−∞∞eiazz−ir dzI_{1} = \displaystyle \int_{-\infty }^{\infty } \frac{e^{iaz}}{z-ir}\, dz; I2=∫−∞∞eiazz+ir dzI_{2} = \displaystyle \int_{-\infty }^{\infty } \frac{e^{iaz}}{z+ir}\, dz; Re(r)>0\mathrm{Re}\left(r\right) > 0; a>0a > 0.

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Problem 5447

Compute the integral ∫0∞xsin⁡axx2+r2 dx\displaystyle \int_{0}^{\infty } \frac{x\sin ax}{x^{2}+r^{2}}\, dx.

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Problem 5448

Compute the integral ∫0∞cos⁡axx2+r2 dx\displaystyle \int_{0}^{\infty } \frac{\cos ax}{x^{2}+r^{2}}\, dx.

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Problem 5449

Compute the integral ∫0∞sin⁡xx dx\displaystyle \int_{0}^{\infty } \frac{\sin x}{x}\, dx.

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Problem 5450

Compute the integral ∫0∞cos⁡ax1+x4 dx\displaystyle \int_{0}^{\infty } \frac{\cos ax}{1+x^{4}}\, dx; a>0a > 0.

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Problem 5451

Compute the integral ∫0∞sin⁡axx(x2+b2) dx\displaystyle \int_{0}^{\infty } \frac{\sin ax}{x(x^{2}+b^{2})}\, dx; a>0a > 0, b>0b > 0.

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Problem 5452

Starting from the identity ∫0∞e−x2 dx=12π\displaystyle \int_{0}^{\infty } e^{-x^{2}}\, dx = \frac{1}{2}\sqrt{\pi }, prove the formulas for the Fresnel integrals:

∫0∞cos⁡(x2) dx=∫0∞sin⁡(x2) dx=12π2. \int _{0}^{\infty } \cos (x^{2})\, dx = \int _{0}^{\infty } \sin (x^{2})\, dx = \frac{1}{2}\sqrt{\frac{\pi }{2}}.
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Problem 5453

Prove the identity: ∫0∞xs−1e−ix dx=Γ(s)e−πsi2\displaystyle \int_{0}^{\infty } x^{s-1}e^{-ix}\, dx = \Gamma (s)e^{-\frac{\pi si}{2}}; 0<Re(s)<10 < \mathrm{Re}\left(s\right) < 1.

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Problem 5454

Prove the identity: ∫0∞cos⁡xn dx=1nΓ(1n)cos⁡π2n\displaystyle \int_{0}^{\infty } \cos x^{n}\, dx = \frac{1}{n}\Gamma \left(\frac{1}{n}\right)\cos \frac{\pi }{2n}; n>1n > 1.

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Problem 5455

Prove the identity: ∫0∞sin⁡xn dx=1nΓ(1n)sin⁡π2n\displaystyle \int_{0}^{\infty } \sin x^{n}\, dx = \frac{1}{n}\Gamma \left(\frac{1}{n}\right)\sin \frac{\pi }{2n}; ∣n∣>1\left|n\right| > 1.

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Problem 5456

Prove the identity: ∫0∞sin⁡xnxn dx=1n−1Γ(1n)cos⁡π2n\displaystyle \int_{0}^{\infty } \frac{\sin x^{n}}{x^{n}}\, dx = \frac{1}{n-1}\Gamma \left(\frac{1}{n}\right)\cos \frac{\pi }{2n}; 0<1n<20 < \frac{1}{n} < 2.

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Problem 5457

Prove the identity 12πi∫a−∞ia+∞iΓ(s)z−s ds=e−z\displaystyle \frac{1}{2\pi i}\int_{a-\infty i}^{a+\infty i} \Gamma (s)z^{-s}\, ds = e^{-z}; a>0a > 0; Re(z)>0\mathrm{Re}\left(z\right) > 0.

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Remark.
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Use the theorem:

lim⁡∣s∣→∞(ln⁡Γ(s)−(s−12)ln⁡s+s−12ln⁡2π)=0, \lim _{\left|s\right|\to \infty }\left(\ln \Gamma (s)-\left(s-\frac{1}{2}\right)\ln s+s-\frac{1}{2}\ln 2\pi \right) = 0,

if ss lies to the right of any given vertical line.

Problem 5458

Prove the identity

12πi∫a−∞ia+∞ixss ds={1,if x>1;0,if x<1;a>0. \frac{1}{2\pi i}\int _{a-\infty i}^{a+\infty i} \frac{x^{s}}{s}\, ds = \begin{cases} 1, & \text{if } x > 1; \\ 0, & \text{if } x < 1; \end{cases} \qquad a > 0.
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Problem 5459

Prove the identity

12πi∫a−∞ia+∞ixss2 ds={ln⁡x,if x>1;0,if x<1;a>0. \frac{1}{2\pi i}\int _{a-\infty i}^{a+\infty i} \frac{x^{s}}{s^{2}}\, ds = \begin{cases} \ln x, & \text{if } x > 1; \\ 0, & \text{if } x < 1; \end{cases} \qquad a > 0.
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Problem 5460

Prove the identity

12πi∫a−∞ia+∞ixs+2s(s+1)(s+2) ds={(x−1)22,if x>1;0,if x<1;a>0. \frac{1}{2\pi i}\int _{a-\infty i}^{a+\infty i} \frac{x^{s+2}}{s(s+1)(s+2)}\, ds = \begin{cases} \dfrac {(x-1)^{2}}{2}, & \text{if } x > 1; \\ 0, & \text{if } x < 1; \end{cases} \qquad a > 0.
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Problem 5461

Prove the identity ∫0π/2f(eiφ)−f(e−iφ)eiφ−e−iφ dφ=π2f(1)−∫01f(ix)+f(−ix)1+x2 dx\displaystyle \int_{0}^{\pi /2} \frac{f(e^{i\varphi })-f(e^{-i\varphi })}{e^{i\varphi }-e^{-i\varphi }}\, d\varphi = \frac{\pi }{2}f(1)-\int_{0}^{1} \frac{f(ix)+f(-ix)}{1+x^{2}}\, dx, if f(x)f(x) is holomorphic in the right half of the disk ∣z∣<1\left|z\right| < 1.

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Problem 5462

Prove the identity ∫0∞cos⁡x−e−xx dx=0\displaystyle \int_{0}^{\infty } \frac{\cos x-e^{-x}}{x}\, dx = 0.

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Problem 5463

Prove the identity ∫0∞x2−a2x2+a2⋅sin⁡xx dx=π(e−a−12)\displaystyle \int_{0}^{\infty } \frac{x^{2}-a^{2}}{x^{2}+a^{2}}\cdot \frac{\sin x}{x}\, dx = \pi \left(e^{-a}-\frac{1}{2}\right); a>0a > 0.

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Problem 5464

Using a contour that consists entirely or in part of the edges of a cut along the positive or negative part of OXOX and circles about the origin of radii ρ→0\rho \to 0 and R→∞R \to \infty, prove the identity: ∫0∞xp−11+x dx=πsin⁡πp\displaystyle \int_{0}^{\infty } \frac{x^{p-1}}{1+x}\, dx = \frac{\pi }{\sin \pi p}; 0<p<10 < p < 1.

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Problem 5465

Using the same technique, prove the identity: v.p.∫0∞xa−11−x dx=πctg⁡aπ\displaystyle \int_{0}^{\infty } \frac{x^{a-1}}{1-x}\, dx = \pi \ctg a\pi; 0<a<10 < a < 1, where

v.p.∫0∞xa−11−x dx=lim⁡σ→+0(∫01−σxa−11−x dx+∫1+σ∞xa−11−x dx). \text{v.p.}\int _{0}^{\infty } \frac{x^{a-1}}{1-x}\, dx = \lim _{\sigma \to +0}\left(\int _{0}^{1-\sigma } \frac{x^{a-1}}{1-x}\, dx+\int _{1+\sigma }^{\infty } \frac{x^{a-1}}{1-x}\, dx\right).
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Problem 5466

Using the same technique, prove the identity:

∫−∞∞eax−ebx1−ex dx=π(ctg⁡aπ−ctg⁡bπ);0<a<1,0<b<1. \int _{-\infty }^{\infty } \frac{e^{ax}-e^{bx}}{1-e^{x}}\, dx = \pi (\ctg a\pi -\ctg b\pi ); \qquad 0 < a < 1, \quad 0 < b < 1.
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Problem 5467

Using the same technique, prove the identity:

∫−∞∞∑λνeaνx1−ex dx=π∑λνctg⁡πaν;0<aν<1,∑λν=0. \int _{-\infty }^{\infty } \frac{\sum \lambda _{\nu }e^{a_{\nu }x}}{1-e^{x}}\, dx = \pi \sum \lambda _{\nu }\ctg \pi a_{\nu }; \qquad 0 < a_{\nu } < 1, \quad \sum \lambda _{\nu } = 0.
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Problem 5468

Using the same technique, prove the identity:

∫0∞x−p1+2xcos⁡λ+x2 dx=πsin⁡πp⋅sin⁡pλsin⁡λ;−1<p<1,−π<λ<π. \int _{0}^{\infty } \frac{x^{-p}}{1+2x\cos \lambda +x^{2}}\, dx = \frac{\pi }{\sin \pi p}\cdot \frac{\sin p\lambda }{\sin \lambda }; \qquad -1 < p < 1, -\pi < \lambda < \pi .
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Problem 5469

Using a contour that includes a cut along the axis OXOX between the points z=0z=0 and z=1z=1, supplemented by circles of radius ρ→0\rho \to 0 about these points, obtain, for −1<p<2-1 < p < 2, the identity:

∫01x1−p(1−x)p(1+x)3 dx=πp(1−p)8sin⁡πp⋅2p. \int _{0}^{1} \frac{x^{1-p}(1-x)^{p}}{(1+x)^{3}}\, dx = \frac{\pi p(1-p)}{8\sin \pi p}\cdot 2^{p}.
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Problem 5470

Using the same technique, obtain, for −1<p<2-1 < p < 2, the identity:

∫01x1−p(1−x)p1+x2 dx=πsin⁡πp(2p2cos⁡πp4−1). \int _{0}^{1} \frac{x^{1-p}(1-x)^{p}}{1+x^{2}}\, dx = \frac{\pi }{\sin \pi p}\left(2^{\frac{p}{2}}\cos \frac{\pi p}{4}-1\right).
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Problem 5471

Using the same technique, obtain, for −1<p<2-1 < p < 2, the identity:

∫01x1−p(1−x)p(a+x)2 dx=πsin⁡πp(1+a−pa(1+aa)p−1−1). \int _{0}^{1} \frac{x^{1-p}(1-x)^{p}}{(a+x)^{2}}\, dx = \frac{\pi }{\sin \pi p}\left(\frac{1+a-p}{a}\left(\frac{1+a}{a}\right)^{p-1}-1\right).
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Problem 5472

Using the same technique, obtain, for −1<p<2-1 < p < 2, the identity:

∫01x1−p(1−x)p1+x dx=πsin⁡πp(1+p−2p). \int _{0}^{1} \frac{x^{1-p}(1-x)^{p}}{1+x}\, dx = \frac{\pi }{\sin \pi p}(1+p-2^{p}).
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Problem 5473

Using the same technique, prove the identity:

∫−11(1+x)1−p(1−x)p1+x2 dx=πsin⁡πp(sin⁡πp2+cos⁡πp2−1). \int _{-1}^{1} \frac{(1+x)^{1-p}(1-x)^{p}}{1+x^{2}}\, dx = \frac{\pi }{\sin \pi p}\left(\sin \frac{\pi p}{2}+\cos \frac{\pi p}{2}-1\right).
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Problem 5474

Prove the identity:

∫0∞dx(x2+a2)(ln⁡2x+π2)=π2a⋅1ln⁡2a+π24−11+a2;a>0. \int _{0}^{\infty } \frac{dx}{(x^{2}+a^{2})(\ln ^{2} x+\pi ^{2})} = \frac{\pi }{2a}\cdot \frac{1}{\ln ^{2} a+\frac{\pi ^{2}}{4}}-\frac{1}{1+a^{2}}; \quad a > 0.
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Problem 5475

Prove the identity:

∫0∞dx(x2+a2)(ln⁡2x+(2n+1)2π2)=12n+1⋅π2a∑k=−nn4k+1(ln⁡2a+(4k+1)2π24)−11+a2. \int _{0}^{\infty } \frac{dx}{(x^{2}+a^{2})(\ln ^{2} x+(2n+1)^{2}\pi ^{2})} = \frac{1}{2n+1}\cdot \frac{\pi }{2a}\sum _{k=-n}^{n} \frac{4k+1}{\left(\ln ^{2} a+(4k+1)^{2}\frac{\pi ^{2}}{4}\right)}-\frac{1}{1+a^{2}}.
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Problem 5476

Prove the identity

∫0∞xa−1φ(x) dx=2πi1−e2aπi∑\Resza−1(φ(z)), \int _{0}^{\infty } x^{a-1}\varphi (x)\, dx = \frac{2\pi i}{1-e^{2a\pi i}}\sum \Res z^{a-1}\left(\varphi (z)\right),

where ∑\Resza−1(φ(z))\sum \Res z^{a-1}\left(\varphi (z)\right) denotes the sum of the residues of the function za−1φ(z)z^{a-1}\varphi (z), taken over all poles of the rational function φ(z)\varphi (z). It is assumed that φ(z)→0\varphi (z) \to 0 as z→∞z \to \infty.

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Problem 5477

Prove the identity

∫0∞φ(x)ln⁡2x+π2 dx=∑\Res(φ(z))ln⁡z−πi, \int _{0}^{\infty } \frac{\varphi (x)}{\ln ^{2} x+\pi ^{2}}\, dx = \sum \Res \frac{\left(\varphi (z)\right)}{\ln z-\pi i},

where φ(z)\varphi (z) is a rational function with no poles on the positive real axis, zφ(z)→0z\varphi (z) \to 0 as z→∞z \to \infty, and ln⁡z\ln z corresponds to arg⁡z\arg z ranging from 00 to 2π2\pi.

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Problem 5478

Prove the theorem: if the series ∑−∞∞f(n)\sum_{-\infty }^{\infty } f(n) converges absolutely, and f(z)f(z) is a meromorphic function, regular at all integer zz, and there exists a sequence of expanding contours C1,C2,…,Cn,…C_{1}, C_{2}, \ldots , C_{n}, \ldots, receding to infinity in every direction as n→∞n \to \infty, such that

∫Cnf(z)ctg⁡πz dz→0,(*) \int _{C_{n}} f(z)\ctg \pi z\, dz \to 0, \tag {*}

then the identity

∑n=−∞∞f(n)=−π∑\Resctg⁡πz(f(z)), \sum _{n=-\infty }^{\infty } f(n) = -\pi \sum \Res \ctg \pi z\left(f(z)\right),

holds, where ∑\Resφ(z)(f(z))\sum \Res \varphi (z)\left(f(z)\right) denotes the sum of the residues of the function φ(z)f(z)\varphi (z)f(z), taken over the poles of f(z)f(z).

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Problem 5479

Under conditions similar to those of the previous problem, where instead of condition (∗)(*) the condition

∫Cnf(z)sin⁡πz dz→0as n→∞, \int _{C_{n}} \frac{f(z)}{\sin \pi z}\, dz \to 0 \quad \text{as } n \to \infty ,

is given, prove that the identity

∑n=−∞∞(−1)nf(n)=−∑\Resπsin⁡πz(f(z)) \sum _{n=-\infty }^{\infty } (-1)^{n}f(n) = -\sum \Res \frac{\pi }{\sin \pi z}\left(f(z)\right)

holds.

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Problem 5480

Prove the identity:

∑n=−∞∞1(a+bn)2=π2b2cosec⁡2πab. \sum _{n=-\infty }^{\infty } \frac{1}{(a+bn)^{2}} = \frac{\pi ^{2}}{b^{2}}\operatorname {cosec}^{2}\frac{\pi a}{b}.
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Problem 5481

Prove the identity:

∑n=1∞1n2+a2=−12a2+π2acth⁡πa. \sum _{n=1}^{\infty } \frac{1}{n^{2}+a^{2}} = -\frac{1}{2a^{2}}+\frac{\pi }{2a}\cth \pi a.
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Problem 5482

Prove the identity:

∑n=1∞(−1)nn2+a2=−12a2+π2a⋅1sh⁡πa. \sum _{n=1}^{\infty } \frac{(-1)^{n}}{n^{2}+a^{2}} = -\frac{1}{2a^{2}}+\frac{\pi }{2a}\cdot \frac{1}{\sh \pi a}.
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Problem 5483

Prove the identity:

∑n=1∞(−1)nnsh⁡πan=−12πa−1a2∑n=1∞(−1)nnsh⁡πna. \sum _{n=1}^{\infty } \frac{(-1)^{n}n}{\sh \pi an} = -\frac{1}{2\pi a}-\frac{1}{a^{2}}\sum _{n=1}^{\infty } \frac{(-1)^{n}n}{\sh \frac{\pi n}{a}}.
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Problem 5484

Prove the identity:

1eπ−e−π−2e2π−e−2π+3e3π−e−3π−…=18π. \frac{1}{e^{\pi }-e^{-\pi }}-\frac{2}{e^{2\pi }-e^{-2\pi }}+\frac{3}{e^{3\pi }-e^{-3\pi }}-\ldots = \frac{1}{8\pi }.
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Problem 5485

Setting, by definition,

f(u)=∫γe−πiz2+2πiuzeπiz−e−πiz dz, f(u) = \int _{\gamma } \frac{e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz,

where the integration proceeds from bottom to top along the entire line x+y=12x+y = \frac{1}{2}, verify the identity:

f(u+1)−f(u)=eπi(u+12)2∫γe−πiz2 dz. f\left(u+1\right)-f(u) = e^{\pi i \left(u+\frac{1}{2}\right)^{2}}\int _{\gamma } e^{-\pi i z^{2}}\, dz.
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Problem 5486

Shifting, in the previous problem, the contour of integration to the line x+y=−12x+y = -\frac{1}{2}, where f(u)=∫γe−πiz2+2πiuzeπiz−e−πiz dzf(u) = \int_{\gamma } \dfrac {e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz (problem 5485), prove the identity:

f(u)=e−2πiuf(u+1)+1. f(u) = e^{-2\pi i u}f(u+1)+1.
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Problem 5487

Prove that ∫γe−πiz2 dz=e3πi4\displaystyle \int_{\gamma } e^{-\pi i z^{2}}\, dz = e^{\frac{3\pi i}{4}} (where γ\gamma is the line x+y=12x+y = \frac{1}{2}, integration from bottom to top, problem 5485), and, using the preceding results, derive Riemann's formula:

∫γe−πiz2+2πiuzeπiz−e−πiz dz=eπiu−eπiu2eπiu−e−πiu. \int _{\gamma } \frac{e^{-\pi i z^{2}+2\pi i uz}}{e^{\pi i z}-e^{-\pi i z}}\, dz = \frac{e^{\pi i u}-e^{\pi i u^{2}}}{e^{\pi i u}-e^{-\pi i u}}.
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Problem 5488

Consider the integral of the function e2πiz2n1−e2πiz\dfrac {e^{\frac{2\pi i z^{2}}{n}}}{1-e^{2\pi i z}}, where nn is a positive integer, along the boundary of the domain

0<x<n2,−ω<y<ω,∣z∣>ρ,∣z−n2∣>π. 0 < x < \frac{n}{2}, \quad -\omega < y < \omega , \quad \left|z\right| > \rho , \quad \left|z-\frac{n}{2}\right| > \pi .

By passing to the limit as ρ→0\rho \to 0, ω→∞\omega \to \infty, obtain the formula for the Gauss sum:

∑ν=0n−1e2πiν2n=1+(−i)n1−i⋅n. \sum _{\nu =0}^{n-1} e^{\frac{2\pi i \nu ^{2}}{n}} = \frac{1+(-i)^{n}}{1-i}\cdot \sqrt{n}.
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Section 14.4Distribution of the Zeros of a Function[23/26%]
Problem 5489

By Rouché's theorem, the functions φ(z)\varphi (z) and ψ(z)\psi (z), regular inside a contour, have the same number of zeros there if on the contour ∣φ(z)−ψ(z)∣<∣φ(z)∣\left|\varphi (z)-\psi (z)\right| < \left|\varphi (z)\right|. Using this, find the number of roots lying in the disk ∣z∣<1\left|z\right| < 1 for the equation z4−5z+1=0z^{4}-5z+1 = 0.

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Problem 5490

By Rouché's theorem, the functions φ(z)\varphi (z) and ψ(z)\psi (z), regular inside a contour, have the same number of zeros there if on the contour ∣φ(z)−ψ(z)∣<∣φ(z)∣\left|\varphi (z)-\psi (z)\right| < \left|\varphi (z)\right|. Using this, find the number of roots lying in the disk ∣z∣<1\left|z\right| < 1 for the equation z6−6z+10=0z^{6}-6z+10 = 0.

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Problem 5491

By Rouché's theorem, the functions φ(z)\varphi (z) and ψ(z)\psi (z), regular inside a contour, have the same number of zeros there if on the contour ∣φ(z)−ψ(z)∣<∣φ(z)∣\left|\varphi (z)-\psi (z)\right| < \left|\varphi (z)\right|. Using this, find the number of roots lying in the disk ∣z∣<1\left|z\right| < 1 for the equation z8−4z5+z2−1=0z^{8}-4z^{5}+z^{2}-1 = 0.

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Problem 5492

By Rouché's theorem, the functions φ(z)\varphi (z) and ψ(z)\psi (z), regular inside a contour, have the same number of zeros there if on the contour ∣φ(z)−ψ(z)∣<∣φ(z)∣\left|\varphi (z)-\psi (z)\right| < \left|\varphi (z)\right|. Using this, find the number of roots lying in the disk ∣z∣<1\left|z\right| < 1 for the equation z=φ(z)z = \varphi (z), where ∣φ(z)∣<1\left|\varphi (z)\right| < 1 for ∣z∣⩽1\left|z\right| \leqslant 1.

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Problem 5493

How many roots does the equation z4+2z3+3z2+z+2=0z^{4}+2z^{3}+3z^{2}+z+2 = 0 have in each square?

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Problem 5494

How many roots does the equation 2z4−3z3+3z2−z+1=02z^{4}-3z^{3}+3z^{2}-z+1 = 0 have in each square?

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Problem 5495

Prove that for λ>1\lambda > 1 the equation z=λ−e−zz = \lambda -e^{-z} has, in the half-plane x>0x > 0, a unique root, and moreover a real one.

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Problem 5496

Prove that z=aezz = ae^{z} has, for 0<a<1e0 < a < \frac{1}{e}, only one root in the disk ∣z∣⩽1\left|z\right| \leqslant 1.

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Problem 5497

Prove that the equation z2=aezz^{2} = ae^{z}, where 0<a<1e0 < a < \frac{1}{e}, has two roots in the disk ∣z∣<1\left|z\right| < 1. They are real and of opposite signs.

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Problem 5498

Prove that the equation 1+z+azn=01+z+az^{n} = 0, where nn is an integer greater than one, has at least one root in the disk ∣z∣⩽2\left|z\right| \leqslant 2 for any aa.

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Problem 5499

Prove that the equation z=tg⁡zz = \tg z has only real roots: a triple root z=0z = 0 and roots zn=(2n+1)π2−ρnz_{n} = (2n+1)\frac{\pi }{2}-\rho_{n}, where ρn>0\rho_{n} > 0, ρn→0\rho_{n} \to 0 as n→∞n \to \infty.

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Problem 5500

Prove that the equation ch⁡zcos⁡z=1\ch z\cos z = 1 has a quadruple root z=0z = 0 and roots of the form ±rn\pm r_{n}, ±irn\pm ir_{n}, rn=(n+12)π+(−1)n−1ρnr_{n} = \left(n+\frac{1}{2}\right)\pi +(-1)^{n-1}\rho_{n}, where ρn>0\rho_{n} > 0 and ρn→0\rho_{n} \to 0 as n→∞n \to \infty.

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Problem 5501

Prove that if the sequence of functions {fn(z)}\left\{ f_{n}(z)\right\}, regular in a domain GG, converges uniformly on compact subsets of this domain to some function f(z)≢0f(z) \not\equiv 0, then every zero of the function f(z)f(z) is the limit of corresponding zeros of the functions fn(z)f_{n}(z).

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Problem 5502

Prove that if, under the conditions of the previous problem (problem 5501), the function f(z)f(z) has inside the domain GG a zero z=z0z = z_{0} of multiplicity mm, then one can find constants ρ0>0\rho_{0} > 0 and C>0C > 0 such that, as soon as the relation ∣fn(z)−f(z)∣<ρ<ρ0\left|f_{n}(z)-f(z)\right| < \rho < \rho_{0} holds on the boundary of the domain GG, the function fn(z)f_{n}(z) has exactly mm zeros in the disk ∣z−z0∣<Cρm\left|z-z_{0}\right| < C\sqrt[m]{\rho }.

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Problem 5503

Prove that if Im(zν)>0\mathrm{Im}\left(z_{\nu }\right) > 0, and ∏ν=1n(z−zν)=φ(z)+iψ(z)\prod_{\nu =1}^{n}(z-z_{\nu }) = \varphi (z)+i\psi (z), where φ(z)\varphi (z) and ψ(z)\psi (z) are polynomials with real coefficients, then the roots of these polynomials are real, simple, and interlacing.

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Problem 5504

Prove that under the same conditions (problem 5503) and for a>0a > 0, the roots of the equations

φ(z)sin⁡az+ψ(z)cos⁡az=0,φ(z)cos⁡az−ψ(z)sin⁡az=0. \varphi (z)\sin az+\psi (z)\cos az = 0, \quad \varphi (z)\cos az-\psi (z)\sin az = 0.

have the same properties.

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Problem 5505

Prove that the number of roots of the real trigonometric polynomial ∑ν=mn(aνcos⁡νφ+bνsin⁡νφ)\sum_{\nu =m}^{n}(a_{\nu }\cos \nu \varphi +b_{\nu }\sin \nu \varphi ), in the interval 0⩽φ<2π0 \leqslant \varphi < 2\pi, is not less than 2m2m and not more than 2n2n.

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Remark.
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Consider the polynomial P(z)=∑ν=mn(aν−ibν)zνP(z) = \sum_{\nu =m}^{n}(a_{\nu }-ib_{\nu })z^{\nu }.

Problem 5506

Prove that for a0<a1<a2<⋯<ana_{0} < a_{1} < a_{2} < \cdots < a_{n} it is easy to prove, by multiplying by 1−z1-z, that the polynomial ∑ν=0naνzν\sum_{\nu =0}^{n} a_{\nu }z^{\nu } has no roots in the region ∣z∣⩾1\left|z\right| \geqslant 1. Using this, show that the trigonometric polynomial ∑ν=0naνcos⁡νφ\sum_{\nu =0}^{n} a_{\nu }\cos \nu \varphi, with the same aνa_{\nu }, has exactly 2n2n distinct roots in the interval 0⩽φ<2π0 \leqslant \varphi < 2\pi (as does the polynomial ∑ν=1naνsin⁡νφ\sum_{\nu =1}^{n} a_{\nu }\sin \nu \varphi).

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Problem 5507

Let, for 0<t<a0 < t < a, the function f(t)f(t) be positive and nondecreasing, and let the integral ∫0af(t) dt\int_{0}^{a} f(t)\, dt make sense. Prove that the entire function f(z)=∫0af(t)cos⁡zt dtf(z) = \int_{0}^{a} f(t)\cos zt\, dt has only real zeros. (Pólya's Theorem.)

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Problem 5508

Prove that under the same conditions (problem 5507) and for b>0b > 0, the entire functions

φ(z)=∫0af(t)cos⁡t(z+b) dt;ψ(z)=∫0af(t)sin⁡z(t+b) dt \varphi (z) = \int _{0}^{a} f(t)\cos t(z+b)\, dt; \quad \psi (z) = \int _{0}^{a} f(t)\sin z(t+b)\, dt

also have only real zeros.

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Problem 5509

Let f(z)f(z) be regular in the disk ∣z∣⩽r\left|z\right| \leqslant r and have, inside the disk, roots a1,a2,…,ana_{1}, a_{2}, \ldots , a_{n}, distinct from zero. By joining each of the zeros aνa_{\nu } with a cut to the circle ∣z∣=r\left|z\right| = r and integrating the corresponding function over the resulting contour, prove Jensen's formula:

ln⁡f(0)+∑ν=1nln⁡r∣aν∣=12π∫02πln⁡∣f(reiφ)∣ dφ. \ln f(0)+\sum _{\nu =1}^{n} \ln \frac{r}{\left|a_{\nu }\right|} = \frac{1}{2\pi }\int _{0}^{2\pi } \ln \left|f(re^{i\varphi })\right|\, d\varphi .
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Problem 5510

Prove that if, under the conditions of the previous problem (problem 5509), the maximum of ∣f(z)∣\left|f(z)\right| equals MM, ∣f(0)∣=m\left|f(0)\right| = m, and NN is the number of zeros of f(z)f(z) in the disk ∣z∣⩽re\left|z\right| \leqslant \frac{r}{e}, then

N⩽ln⁡Mm. N \leqslant \ln \frac{M}{m}.
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Problem 5511

Prove the theorem: the roots of the function

f(z)=Aeax+Bebx+…+Memx, f(z) = Ae^{ax}+Be^{bx}+\ldots +Me^{mx},

where a>b>⋯>ma > b > \cdots > m, and the numbers AA and MM are nonzero, lie in the strip −σ<Re(z)<σ-\sigma < \mathrm{Re}\left(z\right) < \sigma, and NN, the number of them in the rectangle −σ<Re(z)<σ-\sigma < \mathrm{Re}\left(z\right) < \sigma, 0<Im(z)<T0 < \mathrm{Im}\left(z\right) < T, is expressed by the formula

N=a−m2π⋅T+O(1), N = \frac{a-m}{2\pi }\cdot T+O(1),

where O(1)O(1) is bounded as T→∞T \to \infty.

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Section 14.5Expansion of Functions into Partial Fractions[30/0%]
Problem 5512

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

ctg⁡(z−a)+ctg⁡a=∑n=−∞∞(1z−a−nπ+1a+nπ). \ctg (z-a)+\ctg a = \sum _{n=-\infty }^{\infty } \left(\frac{1}{z-a-n\pi }+\frac{1}{a+n\pi }\right).
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Problem 5513

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1sin⁡z=1z+∑n=1∞(−1)n2zz2−n2π2. \frac{1}{\sin z} = \frac{1}{z}+\sum _{n=1}^{\infty } \frac{(-1)^{n}2z}{z^{2}-n^{2}\pi ^{2}}.
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Problem 5514

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1cos⁡z=π∑n=1∞(−1)n(2n−1)z2−(2n−1)2π24. \frac{1}{\cos z} = \pi \sum _{n=1}^{\infty } \frac{(-1)^{n}(2n-1)}{z^{2}-(2n-1)^{2}\frac{\pi ^{2}}{4}}.
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Problem 5515

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

sin⁡zcos⁡2z=∑n=−∞∞(−1)n(z−(2n+1)π2)2. \frac{\sin z}{\cos ^{2} z} = \sum _{n=-\infty }^{\infty } \frac{(-1)^{n}}{\left(z-(2n+1)\frac{\pi }{2}\right)^{2}}.
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Problem 5516

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

π2cos⁡2πz=∑n=−∞∞1(z+12−n)2. \frac{\pi ^{2}}{\cos ^{2} \pi z} = \sum _{n=-\infty }^{\infty } \frac{1}{\left(z+\frac{1}{2}-n\right)^{2}}.
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Problem 5517

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

eazez−1=1z+∑n=1∞2zcos⁡2πna−4πnsin⁡2πnaz2+4n2π2;0<a<1. \frac{e^{az}}{e^{z}-1} = \frac{1}{z}+\sum _{n=1}^{\infty } \frac{2z\cos 2\pi na-4\pi n\sin 2\pi na}{z^{2}+4n^{2}\pi ^{2}}; \quad 0 < a < 1.
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Problem 5518

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

sin⁡azsin⁡πz=2π∑n=1∞(−1)nn⋅sin⁡naz2−n2;−π<a<π. \frac{\sin az}{\sin \pi z} = \frac{2}{\pi }\sum _{n=1}^{\infty } \frac{(-1)^{n}n\cdot \sin na}{z^{2}-n^{2}}; \quad -\pi < a < \pi .
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Problem 5519

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

ch⁡azsh⁡πz=1πz+1π∑n=1∞(−1)n2zcos⁡naz2+n2;−π⩽a⩽π. \frac{\ch az}{\sh \pi z} = \frac{1}{\pi z}+\frac{1}{\pi }\sum _{n=1}^{\infty } \frac{(-1)^{n}2z\cos na}{z^{2}+n^{2}}; \quad -\pi \leqslant a \leqslant \pi .
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Problem 5520

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

1ch⁡2z−cos⁡2z=14z2+4z2∑n=1∞(−1)nnsh⁡πn⋅14z4+n4π4. \frac{1}{\ch 2z-\cos 2z} = \frac{1}{4z^{2}}+4z^{2}\sum _{n=1}^{\infty } \frac{(-1)^{n}n}{\sh \pi n}\cdot \frac{1}{4z^{4}+n^{4}\pi ^{4}}.
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Problem 5521

Prove the validity of the partial-fraction expansion, which holds at all points where the function being expanded is regular:

zsin⁡zsin⁡z−zcos⁡z=3z+∑n=1∞2zz2−λn2;tg⁡λn=λn>0. \frac{z\sin z}{\sin z-z\cos z} = \frac{3}{z}+\sum _{n=1}^{\infty } \frac{2z}{z^{2}-\lambda _{n}^{2}}; \quad \tg \lambda _{n} = \lambda _{n} > 0.
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Problem 5522

Prove the validity of the infinite-product expansion:

cos⁡πz2=∏n=0∞(1−z2(2n+1)2). \cos \frac{\pi z}{2} = \prod _{n=0}^{\infty } \left(1-\frac{z^{2}}{(2n+1)^{2}}\right).
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Problem 5523

Prove the validity of the infinite-product expansion:

sin⁡z−asin⁡a=e−zctg⁡a∏n=−∞∞(1−za+nπ)⋅eza+nπ. \frac{\sin z-a}{\sin a} = e^{-z\ctg a}\prod _{n=-\infty }^{\infty } \left(1-\frac{z}{a+n\pi }\right)\cdot e^{\frac{z}{a+n\pi }}.
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Problem 5524

Prove the validity of the infinite-product expansion:

cos⁡πz−cos⁡πa=−π22(z2−a2)∏n=1∞(1−(z+a2n)2)⋅(1−(z−a2n)2). \cos \pi z-\cos \pi a = -\frac{\pi ^{2}}{2}(z^{2}-a^{2})\prod _{n=1}^{\infty } \left(1-\left(\frac{z+a}{2n}\right)^{2}\right)\cdot \left(1-\left(\frac{z-a}{2n}\right)^{2}\right).
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Problem 5525

Prove the validity of the infinite-product expansion:

ez−ea=ez+a2⋅(z−a)⋅∏n=1∞(1+(z−a2πn)2). e^{z}-e^{a} = e^{\frac{z+a}{2}}\cdot (z-a)\cdot \prod _{n=1}^{\infty } \left(1+\left(\frac{z-a}{2\pi n}\right)^{2}\right).
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Problem 5526

Prove the validity of the infinite-product expansion:

ch⁡z−cos⁡z=z2∏n=1∞(1+z44π4n4). \ch z-\cos z = z^{2}\prod _{n=1}^{\infty } \left(1+\frac{z^{4}}{4\pi ^{4}n^{4}}\right).
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Problem 5527

Prove the validity of the infinite-product expansion:

sin⁡z−zcos⁡z=z23∏n=1∞(1−z2λn2);tg⁡λn=λn>0. \sin z-z\cos z = \frac{z^{2}}{3}\prod _{n=1}^{\infty } \left(1-\frac{z^{2}}{\lambda _{n}^{2}}\right); \quad \tg \lambda _{n} = \lambda _{n} > 0.
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Problem 5528

Prove the validity of the infinite-product expansion:

sin⁡πzπz(1−z)=∏n=1∞(1+z−z2n+n2). \frac{\sin \pi z}{\pi z(1-z)} = \prod _{n=1}^{\infty } \left(1+\frac{z-z^{2}}{n+n^{2}}\right).
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Problem 5529

Prove the validity of the infinite-product expansion:

∏n=1∞(1−zan)=∑n=0∞cnzn;∣a∣>1;∏ν=1n(aν−1)=(−1)ncn. \prod _{n=1}^{\infty } \left(1-\frac{z}{a^{n}}\right) = \sum _{n=0}^{\infty } c_{n}z^{n}; \quad \left|a\right| > 1; \quad \prod _{\nu =1}^{n} (a^{\nu }-1) = \frac{(-1)^{n}}{c_{n}}.
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Problem 5530

Prove the validity of the infinite-product expansion:

∑ν=0∞zν(3ν)!=∏ν=1∞(1+z8rν3). \sum _{\nu =0}^{\infty } \frac{z^{\nu }}{(3\nu )!} = \prod _{\nu =1}^{\infty } \left(1+\frac{z}{8r_{\nu }^{3}}\right). 2cos⁡rν3+e−3rν=0,rν>0. 2\cos r_{\nu }\sqrt{3}+e^{-3r_{\nu }} = 0, \quad r_{\nu } > 0.
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Problem 5531

For Re(s)>0\mathrm{Re}\left(s\right) > 0, Euler's gamma function has the integral representation Γ(s)=∫0∞e−xxs−1 dx\Gamma (s) = \int_{0}^{\infty } e^{-x}x^{s-1}\, dx. It is easy to prove that for 0<x<n0 < x < n the formulas

e−x(1−x2n)<(1−xn)n<e−x e^{-x}\left(1-\frac{x^{2}}{n}\right) < \left(1-\frac{x}{n}\right)^{n} < e^{-x}

hold. Using this, prove Euler's representation of Γ(s)\Gamma (s) as an infinite product:

Γ(s)=lim⁡n→∞1⋅2⋅3⋯n⋅nss(s+1)⋯(s+n). \Gamma (s) = \lim _{n\to \infty } \frac{1\cdot 2\cdot 3\cdots n\cdot n^{s}}{s(s+1)\cdots (s+n)}.
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Problem 5532

Using the preceding representation Γ(s)=lim⁡n→∞1⋅2⋅3⋯n⋅nss(s+1)⋯(s+n)\Gamma (s) = \lim_{n\to \infty } \dfrac {1\cdot 2\cdot 3\cdots n\cdot n^{s}}{s(s+1)\cdots (s+n)} (problem 5531), as well as the formula

lim⁡n→∞(1+12+13+…+1n−ln⁡n)=C, \lim _{n\to \infty } \left(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{n}-\ln n\right) = C,

where CC is Euler's constant, prove the Weierstrass formula

1Γ(s)=seCs∏ν=1∞(1+sν)e−sν. \frac{1}{\Gamma (s)} = se^{Cs}\prod _{\nu =1}^{\infty } \left(1+\frac{s}{\nu }\right)e^{-\frac{s}{\nu }}.
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Problem 5533

Prove the identity:

∏n=1∞n(a+b+n)(a+n)(b+n)=Γ(a+1)Γ(b+1)Γ(a+b+1). \prod _{n=1}^{\infty } \frac{n(a+b+n)}{(a+n)(b+n)} = \frac{\Gamma (a+1)\Gamma (b+1)}{\Gamma (a+b+1)}.
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Problem 5534

Prove the identity:

Γ′(s)Γ(s)=−C−1s+∑n=1∞(1n−1n+s). \frac{\Gamma '(s)}{\Gamma (s)} = -C-\frac{1}{s}+\sum _{n=1}^{\infty } \left(\frac{1}{n}-\frac{1}{n+s}\right).
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Problem 5535

Prove the identity:

(Γ′(s)Γ(s))′=∑ν=0∞1(s+ν)2. \left(\frac{\Gamma '(s)}{\Gamma (s)}\right)' = \sum _{\nu =0}^{\infty } \frac{1}{(s+\nu )^{2}}.
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Problem 5536

Prove the identity:

ln⁡sΓ(s)=(1+12+13+14−C)s+ln⁡1⋅2⋅3⋅4(s+1)(s+2)(s+3)(s+4)+∑ν=2∞(−1)νsννσν, \ln s\Gamma (s) = \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-C\right)s+\ln \frac{1\cdot 2\cdot 3\cdot 4}{(s+1)(s+2)(s+3)(s+4)}+\sum _{\nu =2}^{\infty } \frac{(-1)^{\nu }s^{\nu }}{\nu }\sigma _{\nu },

where σν=15ν+16ν+17ν+…\sigma_{\nu } = \dfrac {1}{5^{\nu }}+\dfrac {1}{6^{\nu }}+\dfrac {1}{7^{\nu }}+\ldots.

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Problem 5537

Prove the identity:

lg⁡10xΓ(x)=lg⁡101⋅2⋅3⋅4(1+x)(2+x)(3+x)(4+x)+0,65409856x−0,00005089x5+ \lg _{10} x\Gamma (x) = \lg _{10} \frac{1\cdot 2\cdot 3\cdot 4}{(1+x)(2+x)(3+x)(4+x)}+0{,}65409856x-0{,}00005089x^{5}+ +0,04805967x2+0,00000740x6−0,00353152x3−0,00000116x7+ +0{,}04805967x^{2}+0{,}00000740x^{6}-0{,}00353152x^{3}-0{,}00000116x^{7}+ +0,00038775x4+0,00000019x8−0,00000003x9… +0{,}00038775x^{4}+0{,}00000019x^{8}-0{,}00000003x^{9}\ldots

Here ∣x∣<5\left|x\right| < 5.

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Problem 5538

From Euler's product for Γ(s)\Gamma (s) it follows that

S=−∑ν=0mln⁡(s+ν)+ln⁡m!+sln⁡m→ln⁡Γ(s). S = -\sum _{\nu =0}^{m} \ln (s+\nu )+\ln m!+s\ln m \to \ln \Gamma (s).

If the factor −1-1 occurring in the sum is replaced using the formula −1=(s+ν−12)−(s+ν+12)-1 = \left(s+\nu -\frac{1}{2}\right)-\left(s+\nu +\frac{1}{2}\right), one easily obtains the identities:

S=∑ν=0m((s+ν+12)ln⁡s+ν+1s+ν−1)+(s−12)ln⁡s−(s+m+12)ln⁡(s+m+1)+ S = \sum _{\nu =0}^{m} \left(\left(s+\nu +\frac{1}{2}\right)\ln \frac{s+\nu +1}{s+\nu }-1\right)+\left(s-\frac{1}{2}\right)\ln s-\left(s+m+\frac{1}{2}\right)\ln (s+m+1)+ +ln⁡m!+sln⁡m+m+1= +\ln m!+s\ln m+m+1 = =∑ν=0m((s+ν+12)ln⁡s+ν+1s+ν−1)+(s−12)ln⁡s−s+m+ln⁡m!−mln⁡m−12ln⁡m+Mm, = \sum _{\nu =0}^{m} \left(\left(s+\nu +\frac{1}{2}\right)\ln \frac{s+\nu +1}{s+\nu }-1\right)+\left(s-\frac{1}{2}\right)\ln s-s+m+\ln m!-m\ln m-\frac{1}{2}\ln m+\frac{M}{m},

where MM is bounded as m→∞m \to \infty. Using this, obtain Gudermann's series:

ln⁡Γ(s)=(s−12)ln⁡s−s+A+∑ν=0∞((s+ν+12)ln⁡s+ν+1s+ν−1), \ln \Gamma (s) = \left(s-\frac{1}{2}\right)\ln s-s+A+\sum _{\nu =0}^{\infty } \left(\left(s+\nu +\frac{1}{2}\right)\ln \frac{s+\nu +1}{s+\nu }-1\right),

where AA is a corresponding constant, and the identity holds for all ss other than the numbers 0,−1,−2,−3,…0, -1, -2, -3, \ldots.

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Problem 5539

Applying the expansion of cos⁡πz2\cos \dfrac {\pi z}{2} as an infinite product (problem 5522), prove the Wallis formula:

π2=lim⁡n→∞(2⋅4⋅6⋯2n3⋅5⋅7⋯(2n+1))2⋅(2n+2), \frac{\pi }{2} = \lim _{n\to \infty } \left(\frac{2\cdot 4\cdot 6\cdots 2n}{3\cdot 5\cdot 7\cdots (2n+1)}\right)^{2}\cdot (2n+2),

and, using it, prove that the constant AA in Gudermann's series (problem 5538) equals ln⁡2π\ln \sqrt{2\pi }.

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Problem 5540

Using Gudermann's series (problem 5538), prove that

ln⁡Γ(s)−((s−12)ln⁡s−s)→ln⁡2π, \ln \Gamma (s)-\left(\left(s-\frac{1}{2}\right)\ln s-s\right) \to \ln \sqrt{2\pi },

if the distance of the point ss from the negative part of the real axis grows without bound, while Re(s)\mathrm{Re}\left(s\right) is bounded below.

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Problem 5541

Prove the validity of the expansion:

Γ(a)Γ(x)Γ(a+x)=1x+∑n=1∞(−1)n(a−1)(a−2)…(a−n)1⋅2⋅3⋯n⋅1x+n;a>0. \frac{\Gamma (a)\Gamma (x)}{\Gamma (a+x)} = \frac{1}{x}+\sum _{n=1}^{\infty } \frac{(-1)^{n}(a-1)(a-2)\ldots (a-n)}{1\cdot 2\cdot 3\cdots n}\cdot \frac{1}{x+n}; \quad a > 0.
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Section 14.6Other Series Expansions[36/53%]
Problem 5542

Expand in increasing powers of ww the root, with respect to zz, of the equation w=2z+z2w = 2z+z^{2} that vanishes at w=0w = 0.

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Remark.
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Some of the problems in this section are solved using the theorem on the Lagrange series. By this theorem, if a point z=az=a lies in a domain SS, and on the contour the inequality ∣tf(z)∣<∣z−a∣\left|tf(z)\right| < \left|z-a\right| holds, then the equation, with respect to zz: z=a+tf(z)z = a+tf(z) has, in this domain, a unique root, representing a function of tt. If F(z)F(z) is a function of this root, regular in SS, then the expansion (Lagrange series) holds:

F(z)=F(a)+∑n=1∞{F′(a)[f(a)]n}(n−1)n!tn. F(z) = F(a)+\sum _{n=1}^{\infty } \frac{\left\{ F'(a)[f(a)]^{n}\right\} ^{(n-1)}}{n!}t^{n}.

In particular, for example:

z=a+∑n=1∞{[f(a)]n}(n−1)n!tn. z = a+\sum _{n=1}^{\infty } \frac{\left\{ [f(a)]^{n}\right\} ^{(n-1)}}{n!}t^{n}.
Problem 5543

Expand in increasing powers of ww the root, with respect to zz, of the equation w=ze−zw = ze^{-z} that vanishes at w=0w = 0.

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Problem 5544

Expand in increasing powers of ww the root, with respect to zz, of the equation (1−z2)w=2z(1-z^{2})w = 2z that vanishes at w=0w = 0.

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Problem 5545

Expand in increasing powers of ww the root, with respect to zz, of the equation (zm+1−b)w=−z(z^{m+1}-b)w = -z that vanishes at w=0w = 0.

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Problem 5546

Expand eze^{z} in increasing powers of ww, if

(z+1)2w=z. (z+1)^{2}w = z.
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Problem 5547

Find the similar expansion (problem 5546) for e−ze^{-z}, where zw=z−azw = z-a.

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Problem 5548

Expand in a series of powers of the eccentricity ee of an elliptical orbit the eccentric anomaly EE, related to the mean anomaly MM by Kepler's equation:

E=M+esin⁡E. E = M+e\sin E.
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Problem 5549

Find the singular points of the equation y3−3y+2x=0y^{3}-3y+2x = 0.

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Remark.
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Algebraic functions satisfy an equation f(x,y)=0f(x,y) = 0, where f(x,y)f(x,y) is a polynomial in the complex variables xx and yy. If both of these variables remain real, then the equation written represents an algebraic curve. If the equation has degree nn in yy, then it defines nn branches y1,y2,…,yny_{1}, y_{2}, \ldots , y_{n} of a multivalued function of xx. They represent regular functions of xx in a domain where their values are finite and unequal. If x0x_{0} is a point lying in such a domain, then each of the functions yνy_{\nu } expands in integer powers of x−x0x-x_{0}. If, however, at the point x0x_{0} several of the yνy_{\nu } coincide, then in a neighborhood of this point the functions yνy_{\nu } expand in fractional powers of x−x0x-x_{0} by the formula

y=∑μ=μ0∞aν(x−x0)μm, y = \sum _{\mu =\mu _{0}}^{\infty } a_{\nu }(x-x_{0})^{\frac{\mu }{m}},

where the number μ0\mu_{0} may be negative, and mm does not exceed nn. The values aνa_{\nu } can be found by substituting these series into the equation. The singular points themselves can be found from the equations:

f(x,y)=0,∂f∂y=0. f(x,y) = 0, \quad \frac{\partial f}{\partial y} = 0.
Problem 5550

Find the singular points of the equation y3−3y2+x6=0y^{3}-3y^{2}+x^{6} = 0.

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Problem 5551

Find the singular points of the equation y3−3y+2x2(2−x2)=0y^{3}-3y+2x^{2}(2-x^{2}) = 0.

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Problem 5552

Find the singular points of the equation y3−3y+2x3(2−x3)=0y^{3}-3y+2x^{3}(2-x^{3}) = 0.

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Problem 5553

Find the first two terms in the expansion of yy in increasing powers of xx for the branches of the curve x3−3xy+y3=0x^{3}-3xy+y^{3} = 0 passing through the origin.

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Problem 5554

Find the first two terms in the expansion of yy in increasing powers of xx for the branches of the curve y4−4y=xy^{4}-4y = x passing through the origin.

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Problem 5555

Find the singular points of the equation y5+ax4−b2xy2=0y^{5}+ax^{4}-b^{2}xy^{2} = 0.

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Problem 5556

Find the singular points of the equation 4(x−1)y5+2xy3−3x3y+x4=04(x-1)y^{5}+2xy^{3}-3x^{3}y+x^{4} = 0.

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Problem 5557

Prove the theorem: a function f(z)f(z), regular inside an ellipse with foci z=±1z = \pm 1 and contour CC, expands within this ellipse in the series:

f(z)=∑n=1∞An(z+z2−1)n−(z−z2−1)nz2−1, f(z) = \sum _{n=1}^{\infty } A_{n}\frac{\left(z+\sqrt{z^{2}-1}\right)^{n}-\left(z-\sqrt{z^{2}-1}\right)^{n}}{\sqrt{z^{2}-1}},

where

An=12πi∫Cf(z)(z−z2−1)n dz;∣z+z2−1∣⩾1. A_{n} = \frac{1}{2\pi i}\int _{C} f(z)\left(z-\sqrt{z^{2}-1}\right)^{n}\, dz; \quad \left|z+\sqrt{z^{2}-1}\right| \geqslant 1.
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Problem 5558

Prove that under the same conditions (problem 5557) the expansion

f(z)=B0+∑n=1∞Bn((z+z2−1)n+(z−z2−1)n), f(z) = B_{0}+\sum _{n=1}^{\infty } B_{n}\left(\left(z+\sqrt{z^{2}-1}\right)^{n}+\left(z-\sqrt{z^{2}-1}\right)^{n}\right),

holds, where

Bn=12πi∫Cf(z)(z−z2−1)nz2−1 dz. B_{n} = \frac{1}{2\pi i}\int _{C} f(z)\frac{\left(z-\sqrt{z^{2}-1}\right)^{n}}{\sqrt{z^{2}-1}}\, dz.
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Problem 5559

Applying the Cauchy expansion (see below) with ψ(z)=eπz+1\psi (z) = e^{\pi z}+1; φ(z)=−1\varphi (z) = -1, obtain the expansion:

f(x)=∑n=0∞2π∫−π/2π/2cos⁡(2n+1)(x−μ)⋅f(μ) dμ;−π2<x<π2. f(x) = \sum _{n=0}^{\infty } \frac{2}{\pi }\int _{-\pi /2}^{\pi /2} \cos (2n+1)(x-\mu )\cdot f(\mu )\, d\mu ; \qquad -\frac{\pi }{2} < x < \frac{\pi }{2}.
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Remark.
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By Cauchy's theorem, for a function f(x)f(x) satisfying the Dirichlet conditions in the interior of the interval a<x<ba < x < b, the expansion

f(x+0)+f(x−0)2=∑n=1∞φ(λn)ψ′(λn)eλnx∫abe−λnμf(μ) dμ, \frac{f(x+0)+f(x-0)}{2} = \sum _{n=1}^{\infty } \frac{\varphi (\lambda _{n})}{\psi '(\lambda _{n})}e^{\lambda _{n}x}\int _{a}^{b} e^{-\lambda _{n}\mu }f(\mu )\, d\mu ,

holds. Here φ(z)\varphi (z) and ψ(z)\psi (z) are entire functions, the zeros of ψ(z)\psi (z) are simple and equal to the numbers λn\lambda_{n}, and for −π2<θ<π2-\frac{\pi }{2} < \theta < \frac{\pi }{2} the identities

lim⁡r→∞(ψ(reiθ)−φ(reiθ)ψ(reiθ)e(x−a)reiθ)=lim⁡r→∞(φ(−reiθ)ψ(−reiθ)e(b−x)reiθ)=0 \lim _{r\to \infty } \left(\frac{\psi (re^{i\theta })-\varphi (re^{i\theta })}{\psi (re^{i\theta })}e^{(x-a)re^{i\theta }}\right) = \lim _{r\to \infty } \left(\frac{\varphi (-re^{i\theta })}{\psi (-re^{i\theta })}e^{(b-x)re^{i\theta }}\right) = 0

hold.

Problem 5560

Using the functions ψ(z)=eaz−z\psi (z) = e^{az}-z; φ(z)=eaz\varphi (z) = e^{az}; a>0a > 0, obtain the expansion:

f(x)=∑n=0∞λnaλn−1eλnx∫−a/2a/2e−λnμf(μ) dμ;−a2<x<a2, f(x) = \sum _{n=0}^{\infty } \frac{\lambda _{n}}{a\lambda _{n}-1}e^{\lambda _{n}x}\int _{-a/2}^{a/2} e^{-\lambda _{n}\mu }f(\mu )\, d\mu ; \qquad -\frac{a}{2} < x < \frac{a}{2},

where λn\lambda_{n} are the roots of the equation z=eazz = e^{az}.

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Problem 5561

Prove the identity, for the same λn\lambda_{n} as before (problem 5560, the roots of the equation z=eazz = e^{az}):

ex=∑n=1∞2λnsh⁡a(1−λn)2(aλn−1)(1−λn)eλnx. e^{x} = \sum _{n=1}^{\infty } \frac{2\lambda _{n}\sh \frac{a(1-\lambda _{n})}{2}}{(a\lambda _{n}-1)(1-\lambda _{n})}e^{\lambda _{n}x}.
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Problem 5562

Prove the identity for the sum of the series of analytic functions:

∑n=1∞zn(1−zn)(1−zn+1)={z(1−z)2for ∣z∣<1;1(1−z)2for ∣z∣>1. \sum _{n=1}^{\infty } \frac{z^{n}}{(1-z^{n})(1-z^{n+1})} = \begin{cases} \dfrac {z}{(1-z)^{2}} & \text{for } \left|z\right| < 1; \\ \dfrac {1}{(1-z)^{2}} & \text{for } \left|z\right| > 1. \end{cases}
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Problem 5563

Prove the identity for the sum of the series of analytic functions:

11−z+∑n=1∞z2n−1z2n−1={1for ∣z∣<1;0for ∣z∣>1. \frac{1}{1-z}+\sum _{n=1}^{\infty } \frac{z^{2^{n-1}}}{z^{2^{n}}-1} = \begin{cases} 1 & \text{for } \left|z\right| < 1; \\ 0 & \text{for } \left|z\right| > 1. \end{cases}
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Problem 5564

Prove the identity for the sum of the series of analytic functions:

1−z+z(z−1)1⋅2−z(z−1)(z−2)1⋅2⋅3+…=0 for Re(z)>0; or=∞ for Re(z)<0. 1-z+\frac{z(z-1)}{1\cdot 2}-\frac{z(z-1)(z-2)}{1\cdot 2\cdot 3}+\ldots = 0 \text{ for } \mathrm{Re}\left(z\right) > 0; \text{ or} = \infty \text{ for } \mathrm{Re}\left(z\right) < 0.
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Problem 5565

Prove the identity for the sum of the series of analytic functions:

1−z21+z2(z2−1)2!2−z2(z2−1)(z2−4)3!2+…=sin⁡πzπz. 1-\frac{z^{2}}{1}+\frac{z^{2}(z^{2}-1)}{2!^{2}}-\frac{z^{2}(z^{2}-1)(z^{2}-4)}{3!^{2}}+\ldots = \frac{\sin \pi z}{\pi z}.
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Problem 5566

Find the domain of convergence of the series ∑n=1∞zn(1+z)(1+z2)…(1+zn)\displaystyle \sum_{n=1}^{\infty } \frac{z^{n}}{(1+z)(1+z^{2})\ldots (1+z^{n})}.

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Problem 5567

Find the domain of convergence of the series ∑n=1∞(−1)nz+n\displaystyle \sum_{n=1}^{\infty } \frac{(-1)^{n}}{z+n}.

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Problem 5568

Find the domain of convergence of the series ∑n=1∞(z(z+n)n)n\displaystyle \sum_{n=1}^{\infty } \left(\frac{z(z+n)}{n}\right)^{n}.

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Problem 5569

Find the domain of convergence of the series ∑n=0∞1n!⋅11+a2nz2\displaystyle \sum_{n=0}^{\infty } \frac{1}{n!}\cdot \frac{1}{1+a^{2n}z^{2}}; ∣a∣>1\left|a\right| > 1.

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Problem 5570

Prove that the function

f(z)=∑m,n=−∞∞1(m+nz)2,m2+n2>0, f(z) = \sum _{m,n=-\infty }^{\infty } \frac{1}{(m+nz)^{2}}, \quad m^{2}+n^{2} > 0,

considered in the upper half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0, is regular there and has the real axis as an essential boundary of continuation.

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Problem 5571

Prove that the function ∑n=1∞1n2⋅1z−e2πni2\displaystyle \sum_{n=1}^{\infty } \frac{1}{n^{2}}\cdot \frac{1}{z-e^{2\pi ni\sqrt{2}}}, regular for ∣z∣<1\left|z\right| < 1, cannot be continued beyond the circle ∣z∣=1\left|z\right| = 1.

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Problem 5572

Prove that the same singularity (problem 5571, the absence of continuation beyond the boundary of the circle of convergence) is exhibited by the series:

∑n=0∞zn!;∑n=0∞12nz2n;∑n=0∞anz2n, \sum _{n=0}^{\infty } z^{n!}; \qquad \sum _{n=0}^{\infty } \frac{1}{2^{n}}z^{2^{n}}; \qquad \sum _{n=0}^{\infty } a_{n}z^{2^{n}},

if ∑n=0∞an\sum_{n=0}^{\infty } a_{n} diverges.

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Problem 5573

Let aa, bb, cc be the numbers represented by the three vertices of a triangle, and xx, yy, zz proper fractions. The series:

∑m=0∞∑n=0∞∑p=0∞xmynzpma+nb+pcm+n+p−ζ,m+n+p>0, \sum _{m=0}^{\infty }\sum _{n=0}^{\infty }\sum _{p=0}^{\infty } \frac{x^{m}y^{n}z^{p}}{\dfrac {ma+nb+pc}{m+n+p}-\zeta }, \quad m+n+p > 0,

represents, outside the triangle, an analytic function of ζ\zeta that cannot be continued inside the triangle. Prove that, contrary to what is stated in Hermite's book Cours d’analyse, the series also converges inside the triangle on an everywhere-dense set of points.

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Problem 5574

Prove the theorem: if a general Dirichlet series converges at σ=σ0\sigma = \sigma_{0}, then it converges for any ss for which σ>σ0\sigma > \sigma_{0}.

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Remark.
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The Abel transform consists in the identity:

∑ν=1naνbν=∑ν=1n−1sν(bν−bν+1)+bnsn, \sum _{\nu =1}^{n} a_{\nu }b_{\nu } = \sum _{\nu =1}^{n-1} s_{\nu }(b_{\nu }-b_{\nu +1})+b_{n}s_{n},

where sν=a1+a2+…+aνs_{\nu } = a_{1}+a_{2}+\ldots +a_{\nu }.

A Dirichlet series is a series of the form ∑n=1∞anns\sum_{n=1}^{\infty } \dfrac {a_{n}}{n^{s}}, where ana_{n} is the coefficient. A generalized Dirichlet series is a series of the form ∑n=1∞anpns\sum_{n=1}^{\infty } \dfrac {a_{n}}{p_{n}^{s}}, where p1,p2,…,pn,…p_{1}, p_{2}, \ldots , p_{n}, \ldots is some infinitely increasing sequence of numbers. Set further s=σ+iτs = \sigma +i\tau. The identity u=O(v)u = O(v) means that the ratio ∣u/v∣\left|u/v\right| remains bounded as vv increases without bound.

For every Dirichlet series ∑n=1∞anns\sum_{n=1}^{\infty } \dfrac {a_{n}}{n^{s}} there exists a number s0s_{0} such that the series converges for s>s0s > s_{0} and diverges for s<s0s < s_{0}. It is called the abscissa of convergence.

Problem 5575

Prove the theorem: if for the quantity An=∑ν=1naνA_{n} = \sum_{\nu =1}^{n} a_{\nu } the identity An=O(nσ0)A_{n} = O(n^{\sigma_{0}}), σ0⩾0\sigma_{0} \geqslant 0, holds, then the Dirichlet series ∑anns\sum \dfrac {a_{n}}{n^{s}} converges for σ>σ0\sigma > \sigma_{0}, and converges absolutely and uniformly for σ>σ0+1+δ\sigma > \sigma_{0}+1+\delta, where δ\delta is any positive constant.

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Problem 5576

Prove the theorem: if the series ∑n=1∞anns\sum_{n=1}^{\infty } \dfrac {a_{n}}{n^{s}} converges at σ=σ0⩾0\sigma = \sigma_{0} \geqslant 0, then An=O(nσ0)A_{n} = O(n^{\sigma_{0}}).

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Problem 5577

Using Pringsheim's theorem, prove that the abscissa of convergence of the general Dirichlet series ∑pn−s\sum p_{n}^{-s} is lim⁡‾ln⁡nln⁡pn\overline{\lim }\dfrac {\ln n}{\ln p_{n}}.

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Remark.
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The upper limit of an infinite sequence of numbers vνv_{\nu } is denoted lim⁡‾vν\overline{\lim } v_{\nu }. It represents the largest of the limit points of the sequence vνv_{\nu }.

Pringsheim's theorem: for a convergent series of positive, decreasing terms ∑uν\sum u_{\nu }, the condition nun→0nu_{n} \to 0 as n→∞n \to \infty must hold.

Section 14.7Generating Functions and Special Polynomials[11/0%]
Problem 5578

Prove the identity:

tetz−1et−1=∑n=1∞φn(z)n!tn. t\frac{e^{tz}-1}{e^{t}-1} = \sum _{n=1}^{\infty } \frac{\varphi _{n}(z)}{n!}t^{n}.

Here φn(z)\varphi_{n}(z) are the Bernoulli polynomials, for which

φn(z+1)−φn(z)=nzn−1. \varphi _{n}(z+1)-\varphi _{n}(z) = nz^{n-1}.

For integer z>0z > 0 we have

1nφn(z)=1n−1+2n−1+…+(z−1)n−1. \frac{1}{n}\varphi _{n}(z) = 1^{n-1}+2^{n-1}+\ldots +(z-1)^{n-1}.
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Remark.
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A function Φ(z,t)\Phi (z,t) is called generating for the functions φn(z)\varphi_{n}(z) if

Φ(z,t)=∑n=0∞φn(z)tn. \Phi (z,t) = \sum _{n=0}^{\infty } \varphi _{n}(z)t^{n}.

If Φ(z,t)\Phi (z,t) is comparatively simple, then from its properties one can derive a range of properties of the functions φn(z)\varphi_{n}(z).

Below are given examples of generating functions for several important classes of polynomials, together with the principal properties of these polynomials that are to be proved using generating functions, as well as by other means.

Problem 5579

Prove that the generating function

e−tz−t22=∑n=0∞Hn(z)tn e^{-tz-\frac{t^{2}}{2}} = \sum _{n=0}^{\infty } H_{n}(z)t^{n}

determines the Chebyshev--Hermite polynomials Hn(z)H_{n}(z), whose principal properties are expressed by the identities:

Hn(z)=1n!ez22dne−z22dzn;Hn′(z)=−Hn−1(z); H_{n}(z) = \frac{1}{n!}e^{\frac{z^{2}}{2}}\frac{d^{n}e^{-\frac{z^{2}}{2}}}{dz^{n}}; \quad H_{n}'(z) = -H_{n-1}(z); nHn(z)+zHn−1(z)+Hn−2(z)=0; nH_{n}(z)+zH_{n-1}(z)+H_{n-2}(z) = 0; Hn′′(z)−zHn′(z)+nHn(z)=0. H_{n}''(z)-zH_{n}'(z)+nH_{n}(z) = 0. ∫−∞∞e−z22Hn(z)Hm(z) dz={0,if m≠n;2πm!,if m=n. \int _{-\infty }^{\infty } e^{-\frac{z^{2}}{2}}H_{n}(z)H_{m}(z)\, dz = \begin{cases} 0, & \text{if } m \neq n; \\ \dfrac {\sqrt{2\pi }}{m!}, & \text{if } m = n. \end{cases}
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Problem 5580

Prove that the generating function

e−zt1−t(1−t)a+1=∑n=0∞Ln(a)(z)tn \frac{e^{-\frac{zt}{1-t}}}{(1-t)^{a+1}} = \sum _{n=0}^{\infty } L_{n}^{(a)}(z)t^{n}

determines the generalized Chebyshev--Laguerre polynomials Ln(a)(z)L_{n}^{(a)}(z) with the properties:

Ln(a)(z)=1n!ezz−adnzn+ae−zdzn, L_{n}^{(a)}(z) = \frac{1}{n!}e^{z}z^{-a}\frac{d^{n}z^{n+a}e^{-z}}{dz^{n}}, nLn(a)(z)+(z−2n−a+1)Ln−1(a)(z)+(n+a−1)Ln−2(a)(z)=0, nL_{n}^{(a)}(z)+(z-2n-a+1)L_{n-1}^{(a)}(z)+(n+a-1)L_{n-2}^{(a)}(z) = 0, z(Ln(a)(z))′′+(a+1−z)(Ln(a)(z))′+nLn(a)(z)=0, z\left(L_{n}^{(a)}(z)\right)''+(a+1-z)\left(L_{n}^{(a)}(z)\right)'+nL_{n}^{(a)}(z) = 0, ∫0∞zae−zLn(a)(z)Lm(a)(z) dz={0for m≠n;Γ(a+n+1)n!for m=n. \int _{0}^{\infty } z^{a}e^{-z}L_{n}^{(a)}(z)L_{m}^{(a)}(z)\, dz = \begin{cases} 0 & \text{for } m \neq n; \\ \dfrac {\Gamma (a+n+1)}{n!} & \text{for } m = n. \end{cases}

For a=0a = 0 these become the ordinary Laguerre polynomials.

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Problem 5581

Prove that the generating function

2α+βr(1−t+r)−α(1+t+r)−β=∑n=0∞Pnα,β(x)tn;r=1−2xt+t2 \frac{2^{\alpha +\beta }}{r}(1-t+r)^{-\alpha }(1+t+r)^{-\beta } = \sum _{n=0}^{\infty } P_{n}^{\alpha ,\beta }(x)t^{n}; \qquad r = \sqrt{1-2xt+t^{2}}

determines the Jacobi polynomials Pnα,βP_{n}^{\alpha ,\beta } with the properties:

(1−x)α(1+x)βPnα,β(x)=(−1)n2n⋅n!dn(1−x)n+α(1+x)n+βdxn. (1-x)^{\alpha }(1+x)^{\beta }P_{n}^{\alpha ,\beta }(x) = \frac{(-1)^{n}}{2^{n}\cdot n!}\frac{d^{n}(1-x)^{n+\alpha }(1+x)^{n+\beta }}{dx^{n}}. ∫−11(1−x)α(1+x)βPnα,β(x)Pmα,β(x) dx=0, if m≠n. \int _{-1}^{1} (1-x)^{\alpha }(1+x)^{\beta }P_{n}^{\alpha ,\beta }(x)P_{m}^{\alpha ,\beta }(x)\, dx = 0, \text{ if } m \neq n. ∫−11(1−x)α(1+x)β(Pnα,β(x))2 dx=2α+β+1α+β+2n+1⋅Γ(α+n+1)Γ(β+n+1)Γ(n+1)Γ(α+β+n+1); \int _{-1}^{1} (1-x)^{\alpha }(1+x)^{\beta }\left(P_{n}^{\alpha ,\beta }(x)\right)^{2}\, dx = \frac{2^{\alpha +\beta +1}}{\alpha +\beta +2n+1}\cdot \frac{\Gamma (\alpha +n+1)\Gamma (\beta +n+1)}{\Gamma (n+1)\Gamma (\alpha +\beta +n+1)}; (n+1)(n+α+β+1)(2n+α+β)Pn+1α,β(x)− (n+1)(n+\alpha +\beta +1)(2n+\alpha +\beta )P_{n+1}^{\alpha ,\beta }(x)- −12(2n+α+β+1)((2n+α+β+2)(2n+α+β)x+(α2−β2))Pnα,β(x)+ -\frac{1}{2}(2n+\alpha +\beta +1)\left((2n+\alpha +\beta +2)(2n+\alpha +\beta )x+(\alpha ^{2}-\beta ^{2})\right)P_{n}^{\alpha ,\beta }(x)+ +(n+α)(n+β)(2n+α+β+2)×Pn−1α,β(x)=0. +(n+\alpha )(n+\beta )(2n+\alpha +\beta +2)\times P_{n-1}^{\alpha ,\beta }(x) = 0. (1−x2)y′′+(β−α−(α+β+2)x)y′+n(n+α+β+1)y=0;y=Pnα,β(x). (1-x^{2})y''+\left(\beta -\alpha -(\alpha +\beta +2)x\right)y'+n(n+\alpha +\beta +1)y = 0; \qquad y = P_{n}^{\alpha ,\beta }(x).
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Problem 5582

Prove that the generating function

11−2tx+t2=∑n=0∞Pn(x)tn \frac{1}{\sqrt{1-2tx+t^{2}}} = \sum _{n=0}^{\infty } P_{n}(x)t^{n}

determines the Legendre polynomials Pn(x)P_{n}(x), which are obtained from the Jacobi polynomials (problem 5581) at α=β=0\alpha = \beta = 0.

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Problem 5583

Prove that the generating function

4−t24−4tx+t2=∑n=0∞Tn(x)tn \frac{4-t^{2}}{4-4tx+t^{2}} = \sum _{n=0}^{\infty } T_{n}(x)t^{n}

determines the Chebyshev polynomials Tn(x)T_{n}(x):

T0(x)=1,Tn(x)=12n−1cos⁡(narccos⁡x)(n=1,2,…). T_{0}(x) = 1, \quad T_{n}(x) = \frac{1}{2^{n-1}}\cos (n\arccos x) \quad (n = 1,2,\ldots ).
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Problem 5584

Prove the identity

11−2tx−t2=∑n=0∞un(x)tn, \frac{1}{1-2tx-t^{2}} = \sum _{n=0}^{\infty } u_{n}(x)t^{n},

where

un(x)=(x+x2+1)n+1−(x−x2+1)n+121+x2. u_{n}(x) = \frac{(x+\sqrt{x^{2}+1})^{n+1}-(x-\sqrt{x^{2}+1})^{n+1}}{2\sqrt{1+x^{2}}}.
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Problem 5585

Using the expansion (1−2xt+t2)−1/2=∑ν=0∞Pν(x)t−ν−1(1-2xt+t^{2})^{-1/2} = \sum_{\nu =0}^{\infty } P_{\nu }(x)t^{-\nu -1}, prove for the Legendre polynomials (problem 5582) the identity:

Pn(cos⁡θ)=12πi∫Ctn1−2tcos⁡θ+t2 dt, P_{n}(\cos \theta ) = \frac{1}{2\pi i}\int _{C} \frac{t^{n}}{\sqrt{1-2t\cos \theta +t^{2}}}\, dt,

where CC is a closed contour inside which lie both singular points of the integrand.

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Problem 5586

From the preceding (problem 5585), obtain the Dirichlet--Mehler formula:

Pn(cos⁡θ)=2π∫0θcos⁡(n+12)t2(cos⁡t−cos⁡θ) dt. P_{n}(\cos \theta ) = \frac{2}{\pi }\int _{0}^{\theta } \frac{\cos \left(n+\frac{1}{2}\right)t}{\sqrt{2(\cos t-\cos \theta )}}\, dt.
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Problem 5587

Using the expansion

(1−2xt+t2)−12=∑ν=0∞Pν(x)t−ν−1, (1-2xt+t^{2})^{-\frac{1}{2}} = \sum _{\nu =0}^{\infty } P_{\nu }(x)t^{-\nu -1},

obtain Laplace's formula (for the Legendre polynomials, problem 5582):

Pn(z)=1π∫−11(z+σz2−1)ndσ1−σ2. P_{n}(z) = \frac{1}{\pi }\int _{-1}^{1} \left(z+\sigma \sqrt{z^{2}-1}\right)^{n}\frac{d\sigma }{\sqrt{1-\sigma ^{2}}}.
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Problem 5588

Applying the generating function (problem 5582) and the identity

1−2tcos⁡θ+t2=(1−teθi)(1−te−θi), 1-2t\cos \theta +t^{2} = (1-te^{\theta i})(1-te^{-\theta i}),

prove the identity:

Pn(cos⁡θ)=1⋅3⋯(2n−1)2⋅4⋯2n× P_{n}(\cos \theta ) = \frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots 2n}\times ×(2cos⁡nθ+12⋅2n2n−1⋅2cos⁡(n−2)θ+1⋅32⋅4⋅2n(2n−2)(2n−1)(2n−3)⋅2cos⁡(n−4)θ+…). \times \left(2\cos n\theta +\frac{1}{2}\cdot \frac{2n}{2n-1}\cdot 2\cos (n-2)\theta +\frac{1\cdot 3}{2\cdot 4}\cdot \frac{2n(2n-2)}{(2n-1)(2n-3)}\cdot 2\cos (n-4)\theta +\ldots \right).
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Section 14.8Conformal Mappings[56/73%]
Problem 5589

What curve does the point zz of the complex plane describe, if z=e(a+bi)tz = e^{(a+bi)t}, where zz is expressed in terms of the real variable tt, ranging from −∞-\infty to ∞\infty, and the coefficients aa, bb are real?

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Problem 5590

What curve does the point zz of the complex plane describe, if z=at+beiωtz = at+be^{i\omega t}, where zz is expressed in terms of the real variable tt, ranging from −∞-\infty to ∞\infty, and the coefficients aa, bb, ω\omega are real?

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Problem 5591

What curve does the point zz of the complex plane describe, if z=aeit+a−1e−itz = ae^{it}+a^{-1}e^{-it}, where zz is expressed in terms of the real variable tt, ranging from −∞-\infty to ∞\infty, and the coefficient aa is real?

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Problem 5592

What curve does the point zz of the complex plane describe, if z=a(1+eit)−2z = a(1+e^{it})^{-2}, where zz is expressed in terms of the real variable tt, ranging from −∞-\infty to ∞\infty, and the coefficient aa is real?

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Problem 5593

What curve does the point zz of the complex plane describe, if z=a(1+it)e−itz = a(1+it)e^{-it}, where zz is expressed in terms of the real variable tt, ranging from −∞-\infty to ∞\infty, and the coefficient aa is real?

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Problem 5594

What line or domain in the plane of the variable zz is defined by the condition ∣z−2∣<∣z∣\left|z-2\right| < \left|z\right|?

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Problem 5595

What line or domain in the plane of the variable zz is defined by the condition ∣z−1∣⩾2∣z−i∣\left|z-1\right| \geqslant 2\left|z-i\right|?

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Problem 5596

What line or domain in the plane of the variable zz is defined by the condition ∣z2−1∣<1\left|z^{2}-1\right| < 1?

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Problem 5597

What line or domain in the plane of the variable zz is defined by the condition Im(z−1)=2\mathrm{Im}\left(z^{-1}\right) = 2?

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Problem 5598

What line or domain in the plane of the variable zz is defined by the condition 0<arg⁡z−iz+i<π40 < \arg \dfrac {z-i}{z+i} < \dfrac {\pi }{4}?

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Problem 5599

What line or domain in the plane of the variable zz is defined by the condition ∣2z∣<∣1+z2∣\left|2z\right| < \left|1+z^{2}\right|?

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Problem 5600

Show that all the roots of the equation

(z−a)n−Kn(z−b)n=0 (z-a)^{n}-K^{n}(z-b)^{n} = 0

lie on a single circle.

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Problem 5601

Show that the function w=z2w = z^{2} maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the first quadrant of the plane zz, onto the upper half-plane ww.

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Problem 5602

Show that the function w=ezw = e^{z} maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the strip 0<y<π20 < y < \dfrac {\pi }{2}, onto the first quadrant of the plane ww.

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Problem 5603

Show that the function w=ln⁡zw = \ln z maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the domain x2+y2=1x^{2}+y^{2} = 1; y>0y > 0, onto the strip u>0u > 0, 0<v<π0 < v < \pi.

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Problem 5604

Show that the function w=cos⁡zw = \cos z maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the strip −π2<x<π2-\dfrac {\pi }{2} < x < \dfrac {\pi }{2}, onto the half-plane v>0v > 0.

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Problem 5605

Show that the function w=1z+iw = \dfrac {1}{z+i} maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the half-plane y⩾0y \geqslant 0, onto the disk u2+v2+v⩽0u^{2}+v^{2}+v \leqslant 0.

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Problem 5606

Show that the function w=i−zi+zw = \dfrac {i-z}{i+z} maps the given domain of the plane z=x+iyz = x+iy conformally and bijectively onto the indicated domain of the plane w=u+ivw = u+iv: the first quadrant of zz, onto the upper half-disk ∣w∣<1\left|w\right| < 1.

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Problem 5607

Into what do the lines ∣z∣=C\left|z\right| = C and arg⁡z=C\arg z = C, and the region ∣z∣>C>1\left|z\right| > C > 1; a<arg⁡z<π−aa < \arg z < \pi -a, where 0<a<π20 < a < \dfrac {\pi }{2}, go under the conformal transformation 2w=z+z−12w = z+z^{-1}?

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Problem 5608

Into what is the upper half-plane zz transformed on the plane ww, if cos⁡w=ch⁡acos⁡z\cos w = \ch a\cos z and at z=π2z = \dfrac {\pi }{2} we have w=π2w = \dfrac {\pi }{2}?

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Problem 5609

The function w=arctg⁡zw = \arctg z transforms the disk ∣z∣<1\left|z\right| < 1 into a strip in the plane ww. Find the width of this strip.

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Problem 5610

Onto what domain is the disk ∣z∣<1\left|z\right| < 1 mapped by the function w=a(nz−zn)w = a(nz-z^{n}), where a>0a > 0, n>1n > 1 an integer? Find the area of this domain.

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Problem 5611

The function w=(1+zn)nw = \left(1+\dfrac {z}{n}\right)^{n} maps some domain SnS_{n} onto the upper half-plane. Find SnS_{n} and lim⁡n→∞Sn\lim_{n\to \infty } S_{n}.

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Problem 5612

The point w=a2zw = \dfrac {a^{2}}{z} is called the reflection of the point zz with respect to the circle ∣z∣=a\left|z\right| = a. Find the reflection, in this circle, of the half-plane Re(z)>2a\mathrm{Re}\left(z\right) > 2a.

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Problem 5613

Prove that under a linear fractional transformation w=az+bcz+dw = \dfrac {az+b}{cz+d}, where ad−bc≠0ad-bc \neq 0, a point and its reflection with respect to a circle go into a point and its reflection with respect to the transformed circle.

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Problem 5614

Prove that the identity

w3−w2w3−w1⋅w−w1w−w2=z3−z2z3−z1⋅z−z1z−z2 \frac{w_{3}-w_{2}}{w_{3}-w_{1}}\cdot \frac{w-w_{1}}{w-w_{2}} = \frac{z_{3}-z_{2}}{z_{3}-z_{1}}\cdot \frac{z-z_{1}}{z-z_{2}}

gives the linear fractional function w(z)w(z) that carries the points z1z_{1}, z2z_{2}, z3z_{3} into the points w1w_{1}, w2w_{2}, w3w_{3}.

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Problem 5615

Prove that the disk ∣z∣<1\left|z\right| < 1 goes into itself under a linear fractional map only for w=eiθ⋅z−a1−a‾zw = e^{i\theta }\cdot \dfrac {z-a}{1-\overline{a}z}, where ∣a∣<1\left|a\right| < 1.

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Problem 5616

Find the conditions under which the linear fractional function w=az+bcz+dw = \dfrac {az+b}{cz+d} carries the disk ∣z∣<1\left|z\right| < 1 into the upper half-plane.

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Problem 5617

Find the function giving the conformal mapping of the disk ∣z∣<1\left|z\right| < 1 onto the half-plane Im(w)<0\mathrm{Im}\left(w\right) < 0 such that the points 11, ii, −i-i go into the points w=+1,0,−1w = +1, 0, -1.

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Problem 5618

Find the function giving the conformal mapping of the same disk (problem 5617) into itself such that the points z=−1,i,1z = -1, i, 1 go into the points w=−1,3i+45,1w = -1, \dfrac {3i+4}{5}, 1.

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Problem 5619

Find the function giving the conformal mapping of the digon bounded by arcs of circles meeting at the points z=az = a, z=bz = b at an angle πα\dfrac {\pi }{\alpha }, onto the half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0 such that the points z=az = a, z=bz = b go into w=0w = 0, w=∞w = \infty.

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Problem 5620

Find the function giving the conformal mapping of the upper half-disk ∣z∣<1\left|z\right| < 1 onto the upper half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0 such that at z=0,1,−1z = 0, 1, -1 we have w=0,1,−1w = 0, 1, -1.

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Problem 5621

Find the function giving the conformal mapping of the quadrant 0<arg⁡z<π20 < \arg z < \dfrac {\pi }{2} onto the disk ∣w∣<1\left|w\right| < 1 such that at z=1+i,0z = 1+i, 0 we have w=0,1w = 0, 1.

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Problem 5622

Find the function giving the conformal mapping of the disk ∣z∣<1\left|z\right| < 1 onto the plane with a cut along the positive part of the real axis.

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Problem 5623

Find the function giving the conformal mapping of the plane with a cut along the whole real axis except the segment between the points aa and bb, onto the region Im(w)>0\mathrm{Im}\left(w\right) > 0.

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Problem 5624

Find the function giving the conformal mapping of the plane with a cut along the segment a<z<ba < z < b onto Im(w)<0\mathrm{Im}\left(w\right) < 0.

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Problem 5625

Find the function giving the conformal mapping of the disk ∣z∣<1\left|z\right| < 1 with a slit along the segment 0<z<10 < z < 1 onto the half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0.

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Problem 5626

Find the function giving the conformal mapping of the strip 0<Re(z)<a0 < \mathrm{Re}\left(z\right) < a onto the half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0.

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Problem 5627

Find the function giving the conformal mapping of the domain ∣z−a∣>a\left|z-a\right| > a; ∣z−b∣<b\left|z-b\right| < b, where 0<a<b0 < a < b, onto the strip 0<Im(w)<π0 < \mathrm{Im}\left(w\right) < \pi.

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Problem 5628

Find the function giving the conformal mapping of the domain 0<a<∣z∣<b0 < a < \left|z\right| < b, 0<arg⁡z<π0 < \arg z < \pi onto a rectangle.

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Problem 5629

Find the function giving the conformal mapping of the domain ∣z−3∣>9\left|z-3\right| > 9; ∣z−8∣<16\left|z-8\right| < 16 onto the annulus between the circles: ρ<∣w∣<1\rho < \left|w\right| < 1.

?
Problem 5630

Find the function giving the conformal mapping of the interior of the ellipse 4x2+5y2=204x^{2}+5y^{2} = 20 with a cut between the foci onto the interior of the annulus 1<∣w∣<A1 < \left|w\right| < A, and find AA.

?
Problem 5631

Find the function giving the conformal mapping of the interior of the parabola y2=4xy^{2} = 4x onto the half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0.

?
Problem 5632

Find the function giving the conformal mapping of the interior of the hyperbola x2sin⁡2α−y2cos⁡2α=sin⁡22αx^{2}\sin^{2}\alpha -y^{2}\cos^{2}\alpha = \sin^{2} 2\alpha onto the half-plane Im(w)>0\mathrm{Im}\left(w\right) > 0.

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Problem 5633

Find the function giving the conformal mapping of the interior of the right half of the lemniscate z2=2a2cos⁡2φz^{2} = 2a^{2}\cos 2\varphi onto the disk ∣w∣⩽1\left|w\right| \leqslant 1.

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Problem 5634

Prove that the function w=∫0zz13−1(1−z)13−1 dzw = \int_{0}^{z} z^{\frac{1}{3}-1}(1-z)^{\frac{1}{3}-1}\, dz maps the upper half-plane zz onto an equilateral triangle with side 32πΓ3(13)\dfrac {\sqrt{3}}{2\pi }\Gamma^{3}\left(\dfrac {1}{3}\right).

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Remark.
?

The following problems are connected with the Christoffel--Schwarz integrals. The half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0 can be mapped onto a polygon by means of the function

w=A∫0z(z−a)α−1(z−b)β−1,…,(z−l)λ−1 dz+B, w = A\int _{0}^{z} (z-a)^{\alpha -1}(z-b)^{\beta -1}, \ldots , (z-l)^{\lambda -1}\, dz+B,

where AA and BB are constants, απ,βπ,…,λπ\alpha \pi , \beta \pi , \ldots , \lambda \pi are the angles of the polygon, and a,b,…,la, b, \ldots , l are points on the axis OXOX.

The same function gives the mapping of the interior of the disk ∣z∣<1\left|z\right| < 1 onto a polygon with the same angles, if the points a,b,…,la, b, \ldots , l lie on the circle ∣z∣=1\left|z\right| = 1. For the polygon to be closed, the condition α+β+…+λ=n−2\alpha +\beta +\ldots +\lambda = n-2 must hold, where nn is the number of quantities a,b,…,la, b, \ldots , l. If the condition is not satisfied, one obtains a mapping onto a region lying on a definite side of the broken line.

Problem 5635

Prove that the function w=∫0zdz1−z4w = \int_{0}^{z} \dfrac {dz}{\sqrt{1-z^{4}}} maps the disk ∣z∣<1\left|z\right| < 1 onto a square with side

14πΓ2(14). \frac{1}{4\sqrt{\pi }}\Gamma ^{2}\left(\frac{1}{4}\right).
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Problem 5636

Prove that the function

w=∫0z(1−z2)−13(1+z2)−23 dz w = \int _{0}^{z} (1-z^{2})^{-\frac{1}{3}}(1+z^{2})^{-\frac{2}{3}}\, dz

maps the interior of the disk ∣z∣<1\left|z\right| < 1 onto a rhombus with angle 60°60° and side 14πΓ(13)Γ(16)\dfrac {1}{4\sqrt{\pi }}\Gamma \left(\dfrac {1}{3}\right)\Gamma \left(\dfrac {1}{6}\right).

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Problem 5637

Prove that the function

w=∫0zdz(1−zn)2n w = \int _{0}^{z} \frac{dz}{(1-z^{n})^{\frac{2}{n}}}

maps the interior of the disk ∣z∣<1\left|z\right| < 1 onto the area of a regular nn-gon, whose side length equals

2πnΓ2(1n)Γ(n−2n)sin⁡2πn. \frac{2}{\pi n}\Gamma ^{2}\left(\frac{1}{n}\right)\Gamma \left(\frac{n-2}{n}\right)\sin ^{2}\frac{\pi }{n}.
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Problem 5638

Onto what domain does the interior of the disk ∣z∣<1\left|z\right| < 1 get mapped by the function (−1<λ<1−2n)\left(-1 < \lambda < 1-\dfrac {2}{n}\right)

w=∫0z(1+zn)λ(1−zn)2n+λ dz? w = \int _{0}^{z} \frac{(1+z^{n})^{\lambda }}{(1-z^{n})^{\frac{2}{n}+\lambda }}\, dz?
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Problem 5639

Find the function giving the conformal mapping of the disk ∣z∣<1\left|z\right| < 1 onto a regular five-pointed star.

?
Problem 5640

Find the function giving the conformal mapping of the same disk (problem 5639) onto the exterior of a regular nn-gon.

?
Problem 5641

Find the function giving the conformal mapping of the half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0 onto the same half-plane with a cut between the points w=0w = 0 and w=iw = i.

?
Problem 5642

Find the function giving the conformal mapping of the half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0 onto the plane with vertical cuts from the points w=nπw = n\pi downward to infinity.

?
Problem 5643

Find the function giving the conformal mapping of the disk ∣z∣<1\left|z\right| < 1 onto the disk ∣w∣<1\left|w\right| < 1 with a radial cut between the points w=12w = \dfrac {1}{2} and w=1w = 1.

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Problem 5644

Find the function giving the conformal mapping of the plane with a cut along the axis OXOX from the point z=1z = 1 to ∞\infty onto the disk ∣w∣<1\left|w\right| < 1.

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Section 14.9The Maximum Modulus Principle[29/0%]
Problem 5645

Applying Cauchy's integral to the function (f(z))n\left(f(z)\right)^{n}, where nn is an integer tending to ∞\infty, prove the theorem: if f(z)f(z) is regular inside a contour CC, and on the contour the inequality ∣f(z)∣⩽M\left|f(z)\right| \leqslant M holds, then it remains valid inside the contour as well.

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Problem 5646

Prove the theorem: if f(z)f(z) is regular inside the disk ∣z∣⩽R\left|z\right| \leqslant R, and ∣f(z)∣<M\left|f(z)\right| < M, while f(0)=0f(0) = 0, then for ∣z∣<R\left|z\right| < R the inequality ∣f(z)∣<MR⋅∣z∣\left|f(z)\right| < \dfrac {M}{R}\cdot \left|z\right| holds, and equality at even one point is possible only when f(z)f(z) has the form f(z)=MReiαzf(z) = \dfrac {M}{R}e^{i\alpha }z. (Schwarz's Lemma.)

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Problem 5647

Prove the theorem: if f(z)f(z) is regular for r<∣z∣<Rr < \left|z\right| < R, then for r<ρ<Rr < \rho < R the identity

12π∫02π∣f(ρeφi)∣2 dφ=∑∣an∣2ρ2n, \frac{1}{2\pi }\int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{2}\, d\varphi = \sum \left|a_{n}\right|^{2}\rho ^{2n},

holds, where ana_{n} is a coefficient of the Laurent series.

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Problem 5648

Prove the theorem: if f(z)=∑n=0∞anznf(z) = \sum_{n=0}^{\infty } a_{n}z^{n} converges for ∣z∣<R\left|z\right| < R, and moreover ∣f(z)∣⩽M\left|f(z)\right| \leqslant M, then ∑∣an∣2R2n⩽M2\sum \left|a_{n}\right|^{2}R^{2n} \leqslant M^{2}.

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Problem 5649

Prove that if the identity w=∑n=0∞anznw = \sum_{n=0}^{\infty } a_{n}z^{n} gives a mapping of the disk ∣z∣<R\left|z\right| < R onto an area SS, then S=π∑n=0∞n∣an∣2R2nS = \pi \sum_{n=0}^{\infty } n\left|a_{n}\right|^{2}R^{2n}.

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Problem 5650

Prove the theorem: if f(z)f(z) is regular in the disk ∣z∣<R\left|z\right| < R and is not constant, then the function

I(r)=∫02π∣f(reφi)∣2 dφ I(r) = \int _{0}^{2\pi } \left|f(re^{\varphi i})\right|^{2}\, d\varphi

increases together with rr for 0<r<R0 < r < R.

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Problem 5651

Prove the theorem: among all functions f(z)f(z), regular inside the disk ∣z∣<R\left|z\right| < R and satisfying the conditions ∫02π∣f(Reφi)∣2 dφ=M\int_{0}^{2\pi } \left|f(Re^{\varphi i})\right|^{2}\, d\varphi = M, the linear function gives the smallest area onto which the disk ∣z∣<R\left|z\right| < R is mapped.

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Problem 5652

Prove the theorem: for a function f(z)f(z), regular for r<∣z∣<Rr < \left|z\right| < R, the quantity

I(ρ)=∫02π∣f(ρeφi)∣2 dφ=K(ln⁡ρ) I(\rho ) = \int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{2}\, d\varphi = K(\ln \rho )

is a convex function of ln⁡ρ\ln \rho, i.e., the inequality holds (for m>0m > 0, n>0n > 0):

K(mln⁡ρ1+nln⁡ρ2m+n)⩽mK(ln⁡ρ1)+nK(ln⁡ρ2)m+n. K\left(\frac{m\ln \rho _{1}+n\ln \rho _{2}}{m+n}\right) \leqslant \frac{mK(\ln \rho _{1})+nK(\ln \rho _{2})}{m+n}.
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Problem 5653

Prove the theorem: if f(z)f(z), besides the previous conditions (problem 5652), is also nonzero for r<∣z∣<Rr < \left|z\right| < R, then the quantity

Iσ(ρ)=∫02π∣f(ρeφi)∣σ dφ I_{\sigma }(\rho ) = \int _{0}^{2\pi } \left|f(\rho e^{\varphi i})\right|^{\sigma }\, d\varphi

is likewise, for any σ>0\sigma > 0, a convex function of ln⁡ρ\ln \rho.

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Problem 5654

Show that the condition f(z)≠0f(z) \neq 0 in the previous problem (problem 5653) can be dispensed with.

?
Problem 5655

Prove the theorem: if f(z)f(z) is regular for ∣z∣<R\left|z\right| < R, then the quantity Iσ(ρ)I_{\sigma }(\rho ) (problem 5653) increases together with ρ\rho.

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Problem 5656

Prove the theorem: if f(z)f(z) is regular for r<∣z∣<Rr < \left|z\right| < R, and M(ρ)M(\rho ) is the maximum modulus of f(ρeφi)f(\rho e^{\varphi i}), then for r<ρ<Rr < \rho < R the quantity M(ρ)M(\rho ) is a convex function of ln⁡ρ\ln \rho (Hadamard's Three-Circles Theorem).

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Remark.
?

Hint. It is useful to consider the function znf(z)=F(z)z^{n}f(z) = F(z), choosing nn so that the quantities M(r)M(r) and M(R)M(R) for F(z)F(z) are equal.

Problem 5657

Prove the theorem: if a polynomial f(z)f(z) of degree nn does not exceed MM in modulus on the interval −1⩽z⩽1-1 \leqslant z \leqslant 1, then for any zz the inequality ∣f(z)∣⩽M(a+b)n\left|f(z)\right| \leqslant M(a+b)^{n} holds, where aa and bb are the semi-axes of the ellipse passing through the point zz and having its foci at the points z=±1z = \pm 1.

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Remark.
?

Hint. Set z=12(ζ+1ζ)z = \dfrac {1}{2}\left(\zeta +\dfrac {1}{\zeta }\right) and apply the theorem on the modulus to the function

ζ−nf(ζ2+12ζ). \zeta ^{-n}f\left(\frac{\zeta ^{2}+1}{2\zeta }\right).
Problem 5658

Prove the theorem: if for ∣z∣<1\left|z\right| < 1 the function f(z)f(z) is regular and ∣f(z)∣<M\left|f(z)\right| < M, while f(α)=0f(\alpha ) = 0, where ∣α∣<1\left|\alpha \right| < 1, then for ∣z∣<1\left|z\right| < 1 the inequality

∣f(z)∣⩽M⋅∣z−α1−α‾z∣ \left|f(z)\right| \leqslant M\cdot \left|\frac{z-\alpha }{1-\overline{\alpha }z}\right|

holds.

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Problem 5659

Prove the theorem: if f(z)f(z) is regular for ∣z∣<1\left|z\right| < 1; f(0)≠0f(0) \neq 0; ∣f(z)∣<M\left|f(z)\right| < M, and f(z)=0f(z) = 0 at z=z1,z2,z3,…z = z_{1}, z_{2}, z_{3}, \ldots, then the infinite product ∏ν=1∞∣zν∣\prod_{\nu =1}^{\infty } \left|z_{\nu }\right| converges.

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Remark.
?

Hint. Consider the inequality

∣f(z)∣⩽M∣z−z11−z1‾z⋅z−z21−z2‾z⋯z−zn1−zn‾z∣. \left|f(z)\right| \leqslant M\left|\frac{z-z_{1}}{1-\overline{z_{1}}z}\cdot \frac{z-z_{2}}{1-\overline{z_{2}}z}\cdots \frac{z-z_{n}}{1-\overline{z_{n}}z}\right|.
Problem 5660

Prove the theorem: if for ∣z∣<1\left|z\right| < 1 the function f(z)f(z) is regular and bounded, and for a<φ<ba < \varphi < b the quantity f(reφi)f(re^{\varphi i}) tends uniformly to zero as r→1r \to 1, then f(z)f(z) is identically zero.

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Remark.
?

Hint. Consider the function f(z)⋅f(ωz)…f(ωn−1z)f(z)\cdot f(\omega z)\ldots f(\omega^{n-1}z), where ω=e2πi/n\omega = e^{2\pi i/n}, and nn is sufficiently large.

Problem 5661

Prove the theorem: if the sequence fn(z)f_{n}(z), for ∣z∣⩽1\left|z\right| \leqslant 1, is uniformly bounded and tends to 00 as n→∞n \to \infty on a set of points having the origin as a limit point, then fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty at any interior point of the disk ∣z∣<1\left|z\right| < 1.

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Remark.
?

Hint. Let z1,z2,…,zmz_{1}, z_{2}, \ldots , z_{m} be points lying inside a fixed disk ∣z∣>ρ\left|z\right| > \rho, at which fn(z)→0f_{n}(z) \to 0. Considering the formula

fn(z)=∑ν=1nfn(zν)ω′(zν)⋅ω(z)z−zν+ω(z)2πi∫∣t∣=1fn(t)ω(t)(t−z) dt;ω(z)=∏ν=1m(z−zν), f_{n}(z) = \sum _{\nu =1}^{n} \frac{f_{n}(z_{\nu })}{\omega '(z_{\nu })}\cdot \frac{\omega (z)}{z-z_{\nu }}+\frac{\omega (z)}{2\pi i}\int _{\left|t\right|=1} \frac{f_{n}(t)}{\omega (t)(t-z)}\, dt; \qquad \omega (z) = \prod _{\nu =1}^{m} (z-z_{\nu }),

fix, for a given ε>0\varepsilon > 0, a sufficiently small ρ\rho, then a correspondingly large mm, and then let nn tend to ∞\infty.

Problem 5662

Prove the theorem: if the functions fn(z)f_{n}(z) are regular inside a contour CC and uniformly bounded on it, and inside the contour there is a limit point of points at which fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty, then fn(z)→0f_{n}(z) \to 0 at every point inside the contour as well. (Vitali's Theorem.)

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Problem 5663

Prove the theorem: if the functions fn(z)f_{n}(z) are regular in the disk ∣z∣<1\left|z\right| < 1, have no roots there, and are bounded in modulus by 11, and, moreover, fn(0)→0f_{n}(0) \to 0 as n→∞n \to \infty, then throughout the interior of the disk ∣z∣<1\left|z\right| < 1 the functions fn(z)→0f_{n}(z) \to 0 as n→∞n \to \infty.

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Remark.
?

Hint. Use the Poisson integral for ln⁡∣fn(z)∣\ln \left|f_{n}(z)\right|.

Problem 5664

Let zνz_{\nu } be the roots of an entire function f(z)f(z), and

0<∣z1∣⩽∣z2∣⩽∣z3∣⩽⋯⩽∣zN−1∣⩽R<∣zN∣, 0 < \left|z_{1}\right| \leqslant \left|z_{2}\right| \leqslant \left|z_{3}\right| \leqslant \cdots \leqslant \left|z_{N-1}\right| \leqslant R < \left|z_{N}\right|,

while for the function f(z)f(z) the inequality

∣f(z)∣<Ce∣z∣σ \left|f(z)\right| < Ce^{\left|z\right|^{\sigma }}

holds. Prove that for ∣z∣⩽3R\left|z\right| \leqslant 3R

∣f(0)∣<∣f(z)∏ν=1N−1(1−zzν)∣<Ce(3R)σ2N−1. \left|f(0)\right| < \left|\frac{f(z)}{\prod _{\nu =1}^{N-1} \left(1-\dfrac {z}{z_{\nu }}\right)}\right| < \frac{Ce^{(3R)^{\sigma }}}{2^{N-1}}.
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Problem 5665

Prove that, under the same conditions (problem 5664), the series ∑ν=1∞1∣zν∣n\displaystyle \sum_{\nu =1}^{\infty } \frac{1}{\left|z_{\nu }\right|^{n}} converges for n>σn > \sigma.

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Problem 5666

Under the same conditions (problem 5664), prove the Weierstrass infinite-product expansion:

f(z)=eg(z)∏ν=1∞φν(z), f(z) = e^{g(z)}\prod _{\nu =1}^{\infty } \varphi _{\nu }(z),

where g(z)g(z) is some entire function, and

φν(z)=(1−zzν)ezzν+z22zν2+…+zn−1(n−1)zνn−1. \varphi _{\nu }(z) = \left(1-\frac{z}{z_{\nu }}\right)e^{\frac{z}{z_{\nu }}+\frac{z^{2}}{2z_{\nu }^{2}}+\ldots +\frac{z^{n-1}}{(n-1)z_{\nu }^{n-1}}}.
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Problem 5667

Let

f(z)=∑n=0∞anzn f(z) = \sum _{n=0}^{\infty } a_{n}z^{n}

satisfy, in the disk ∣z∣⩽R\left|z\right| \leqslant R, the condition: Re(f(z))⩽M\mathrm{Re}\left(f(z)\right) \leqslant M, where M>0M > 0. Then

∣a1∣<2∣M−Re(a0)∣R. \left|a_{1}\right| < \frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|}{R}.
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Remark.
?

Hint. Consider first the case a0=0a_{0} = 0, using the expansion f(z)2M−f(z)=a12Mz+bz2+⋯\dfrac {f(z)}{2M-f(z)} = \dfrac {a_{1}}{2M}z+bz^{2}+\cdots and the relation ∣f(z)∣<∣2M−f(z)∣\left|f(z)\right| < \left|2M-f(z)\right|.

Problem 5668

Considering, under the same conditions (problem 5667), the function

F(z)=1n∑ν=1n(f(ωνz)−a0), F(z) = \frac{1}{n}\sum _{\nu =1}^{n} \left(f(\omega ^{\nu }z)-a_{0}\right),

where ω=e2πin\omega = e^{\frac{2\pi i}{n}}, prove the inequality:

∣an∣⩽2∣M−Re(a0)∣Rn. \left|a_{n}\right| \leqslant \frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|}{R^{n}}.
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Problem 5669

Under the same condition on the growth of f(z)f(z) as before (see problem 5664), prove that for ∣z∣⩽2R\left|z\right| \leqslant 2R, and hence also for ∣z∣⩽R\left|z\right| \leqslant R, we have:

∣ψ(z)∣=∣f(z)∏ν=1N(1−zzν)∣<Ce(2R)σ, \left|\psi (z)\right| = \left|\frac{f(z)}{\prod _{\nu =1}^{N} \left(1-\dfrac {z}{z_{\nu }}\right)}\right| < Ce^{(2R)^{\sigma }},

where in the product ∣zν∣⩽R\left|z_{\nu }\right| \leqslant R.

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Problem 5670

Using the fact that ln⁡ψ(z)\ln \psi (z) is a regular function of zz for ∣z∣⩽R\left|z\right| \leqslant R (problem 5669), prove that at z=0z = 0 the quantity

(ln⁡f(z))(n)−∑ν=1N(ln⁡φν(z))(n) \left(\ln f(z)\right)^{(n)}-\sum _{\nu =1}^{N} \left(\ln \varphi _{\nu }(z)\right)^{(n)}

is, in absolute value, less than cRn−σ\dfrac {c}{R^{n-\sigma }}, where cc is a corresponding constant.

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Problem 5671

Prove Hadamard's theorem: if f(z)f(z) satisfies, for every zz, the condition ∣f(z)∣<Ce∣z∣σ\left|f(z)\right| < Ce^{\left|z\right|^{\sigma }}, where σ<n\sigma < n and f(0)≠0f(0) \neq 0, then the Weierstrass expansion

f(z)=eg(z)⋅∏ν=1∞φν(z), f(z) = e^{g(z)}\cdot \prod _{\nu =1}^{\infty } \varphi _{\nu }(z),

holds, where g(z)g(z) is a polynomial whose degree is less than nn.

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Problem 5672

Prove Carathéodory's theorem: if f(z)=∑n=0∞an(z−a)nf(z) = \sum_{n=0}^{\infty } a_{n}(z-a)^{n} is regular in the disk ∣z−a∣⩽R\left|z-a\right| \leqslant R and there Re(f(z))<M\mathrm{Re}\left(f(z)\right) < M, where M>0M > 0, then for ∣z−a∣=r<R\left|z-a\right| = r < R the inequality

∣f(z)∣<∣a0∣+2∣M−Re(a0)∣⋅rR−r \left|f(z)\right| < \left|a_{0}\right|+\frac{2\left|M-\mathrm{Re}\left(a_{0}\right)\right|\cdot r}{R-r}

holds.

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Problem 5673

Prove Lindelöf's theorem: if ∣f(z)∣⩽C\left|f(z)\right| \leqslant C on the edges of the region −a<x<a-a < x < a, and inside this region ∣f(z)∣<eλ∣y∣\left|f(z)\right| < e^{\lambda \left|y\right|}, where λ\lambda is some constant, then also inside the region ∣f(z)∣⩽C\left|f(z)\right| \leqslant C, and moreover equality at an interior point is possible only when f(z)≡Cf(z) \equiv C.

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Remark.
?

Hint. It is useful to study the function f(z)eεz2f(z)e^{\varepsilon z^{2}} for ε>0\varepsilon > 0.

Section 14.10Differential Equations in a Complex Variable[46/46%]
Problem 5674

Solve, by Cauchy's method, the equation y′′−2y′+y=f(x)y''-2y'+y = f(x); y0=ay_{0} = a, y0′=by_{0}' = b, given these initial conditions.

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Remark.
?

Cauchy's method for finding the integrals of a linear system of differential equations with constant coefficients

dxmdt+∑ν=1nαmνxν=φm(t),m=1,2,…,n(*) \frac{dx_{m}}{dt}+\sum _{\nu =1}^{n} \alpha _{m\nu }x_{\nu } = \varphi _{m}(t), \quad m = 1,2,\ldots ,n \tag {*}

consists in the following:

  1. Introducing the symbol δμν=0\delta_{\mu \nu } = 0 for μ≠ν\mu \neq \nu and δνν=1\delta_{\nu \nu } = 1, we solve the equations

    ∑ν=1n(αμν+sδμν)Aν=ξμ;μ=1,2,…,n;(**) \sum _{\nu =1}^{n} (\alpha _{\mu \nu }+s\delta _{\mu \nu })A_{\nu } = \xi _{\mu }; \quad \mu = 1,2,\ldots ,n; \tag {**}

    which gives:

    Aν=∑μ=1nBνμ(s)ξμΔ(s), A_{\nu } = \frac{\sum _{\mu =1}^{n} B_{\nu \mu }(s)\xi _{\mu }}{\Delta (s)},

    where Δ(s)\Delta (s) is the determinant of the system (∗∗)(**).

  2. We form the functions

    ωm(s)=∑μ=1nBmμ(s)(ξμ+∫0tφμ(τ)e−sτ dτ)esτ. \omega _{m}(s) = \sum _{\mu =1}^{n} B_{m\mu }(s)\left(\xi _{\mu }+\int _{0}^{t} \varphi _{\mu }(\tau )e^{-s\tau }\, d\tau \right)e^{s\tau }.

Then the equations (∗)(*) have the system of particular integrals:

xm=12πi∫Cωm(s)Δ(s) ds,m=1,2,…,n. x_{m} = \frac{1}{2\pi i}\int _{C} \frac{\omega _{m}(s)}{\Delta (s)}\, ds, \quad m = 1,2,\ldots ,n.

Here the integration is taken in the positive direction around the circle ∣s∣=R\left|s\right| = R, large enough that all the roots of the equation Δ(s)=0\Delta (s) = 0 lie inside the circle. At t=0t = 0 the quantities xmx_{m} reduce to ξm\xi_{m}.

For a single equation

y(n)+a1y(n−1)+…+any=f(x) y^{(n)}+a_{1}y^{(n-1)}+\ldots +a_{n}y = f(x)

Cauchy's rule becomes the following:

  1. We form the function

    φ(s)=∑ν=0n−1y0(ν)ψν(s), \varphi (s) = \sum _{\nu =0}^{n-1} y_{0}^{(\nu )}\psi _{\nu }(s),

    where ψν(s)=sn−ν−1+a1sn−ν−2+…+an−ν−1\psi_{\nu }(s) = s^{n-\nu -1}+a_{1}s^{n-\nu -2}+\ldots +a_{n-\nu -1}.

  2. Then

    y=12πi∫∑ν=0n−1y0(ν)ψν(s)⋅esx+∫0xe(x−t)sf(t) dtsn+a1sn−1+…+an ds. y = \frac{1}{2\pi i}\int \frac{\sum _{\nu =0}^{n-1} y_{0}^{(\nu )}\psi _{\nu }(s)\cdot e^{sx}+\int _{0}^{x} e^{(x-t)s}f(t)\, dt}{s^{n}+a_{1}s^{n-1}+\ldots +a_{n}}\, ds.

Here y0,y0′,…,y0(n−1)y_{0}, y_{0}', \ldots , y_{0}^{(n-1)} are the values at x=0x = 0, and the integration proceeds around the circle ∣s∣=R\left|s\right| = R, inside which lie all the roots of the denominator.

Problem 5675

Solve, by Cauchy's method (problem 5674), the system of equations x′−x+y=0x'-x+y = 0, y′+x+z=0y'+x+z = 0, z′−x+y=0z'-x+y = 0; x0=ax_{0} = a, y0=by_{0} = b, z0=cz_{0} = c, given these initial conditions.

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Problem 5676

Solve, by Cauchy's method (problem 5674), the system of equations x′+x+y=etx'+x+y = e^{t}, y′−x+3y=e−ty'-x+3y = e^{-t}; x0=−y0=1x_{0} = -y_{0} = 1, given these initial conditions.

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Problem 5677

Solve, by Cauchy's method (problem 5674), the system of equations x′=4x+2y+2zx' = 4x+2y+2z, y′+x+z=0y'+x+z = 0, z′+2x+y=0z'+2x+y = 0; −x0=y0=z0=1-x_{0} = y_{0} = z_{0} = 1, given these initial conditions.

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Problem 5678

Solve, by Cauchy's method (problem 5674), the system of equations x′′=3x+4yx'' = 3x+4y, y′′+x+y=0y''+x+y = 0; x0=y0′=1x_{0} = y_{0}' = 1, y0=x0′=0y_{0} = x_{0}' = 0, given these initial conditions.

?
Problem 5679

Find the general integral by means of power series for the equation y′′−xy=0y''-xy = 0.

?
Problem 5680

Find the general integral by means of power series for the equation y′′−x2y=0y''-x^{2}y = 0.

?
Problem 5681

Find the general integral by means of power series for the equation y′′′+xy=0y'''+xy = 0.

?
Problem 5682

Find the general integral by means of power series for the equation y′′′+x3y=0y'''+x^{3}y = 0.

?
Problem 5683

Find the general integral by means of power series for the equation (1−x2)y′′−4xy′−2y=0(1-x^{2})y''-4xy'-2y = 0.

?
Problem 5684

Find the general integral by means of power series for the equation xy′′+y′+y=0xy''+y'+y = 0.

?
Problem 5685

Find the general integral by means of power series for the equation xy′′+2y′+xy=0xy''+2y'+xy = 0.

?
Problem 5686

Prove that the equation (1−x2)y′′−5xy′−4y=0(1-x^{2})y''-5xy'-4y = 0 is satisfied by the series

∑n=0∞(n!)2(2n)!(2x)2nand∑n=0∞(2n+1)!(n!)2⋅(x2)2n+1, \sum _{n=0}^{\infty } \frac{(n!)^{2}}{(2n)!}(2x)^{2n} \quad \text{and} \quad \sum _{n=0}^{\infty } \frac{(2n+1)!}{(n!)^{2}}\cdot \left(\frac{x}{2}\right)^{2n+1},

the first of which equals 12((arcsin⁡x)2)′′\dfrac {1}{2}\left((\arcsin x)^{2}\right)''.

?
Problem 5687

Prove that the general integral of the equation

(1−x2)y′′−(α+β+1)xy′−αβy=0 (1-x^{2})y''-(\alpha +\beta +1)xy'-\alpha \beta y = 0

equals C1y1+C2y2C_{1}y_{1}+C_{2}y_{2}, where

y1=1+∑n=1∞α(α+2)⋯(α+2n−2)β(β+2)⋯(β+2n−2)(2n)!x2n, y_{1} = 1+\sum _{n=1}^{\infty } \frac{\alpha (\alpha +2)\cdots (\alpha +2n-2)\beta (\beta +2)\cdots (\beta +2n-2)}{(2n)!}x^{2n}, y2=∑n=0∞(1(2n−1)!(α+1)(α+3)⋯(α+2n−1)(β+1)(β+3)⋯(β+2n−1)x2n+1). y_{2} = \sum _{n=0}^{\infty } \left(\frac{1}{(2n-1)!}(\alpha +1)(\alpha +3)\cdots (\alpha +2n-1)(\beta +1)(\beta +3)\cdots (\beta +2n-1)x^{2n+1}\right).
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Problem 5688

Prove that the equation x2y′+(x−1)y+1=0x^{2}y'+(x-1)y+1 = 0, whose general integral is

y=Cxe−1x−1xe−1x∫xe1x dx, y = \frac{C}{x}e^{-\frac{1}{x}}-\frac{1}{x}e^{-\frac{1}{x}}\int xe^{\frac{1}{x}}\, dx,

is formally satisfied by the series ∑n=0∞n!xn\sum_{n=0}^{\infty } n!x^{n}, which diverges for every x≠0x \neq 0.

?
Problem 5689

Find the integral of the equation

(1−x2)y′′−2xy′+n(n+1)y=0, (1-x^{2})y''-2xy'+n(n+1)y = 0,

where 2n2n is a non-integer, as a series in positive powers of xx.

?
Problem 5690

Find the integral of the equation (1−x2)y′′−2xy′+n(n+1)y=0(1-x^{2})y''-2xy'+n(n+1)y = 0, where 2n2n is a non-integer (problem 5689), as a series in negative powers of xx.

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Problem 5691

Find the general integral of the equation

4x(x−1)y′′+4((a+2)x−a−1)y′+(2a+1)y=0, 4x(x-1)y''+4\left((a+2)x-a-1\right)y'+(2a+1)y = 0,

representing it by the identity y=C1φ(x,a)+C2x−aφ(−a,x)y = C_{1}\varphi (x,a)+C_{2}x^{-a}\varphi (-a,x), where φ(a,x)\varphi (a,x) is a series in powers of xx, equal to 11 at x=0x = 0.

?
Problem 5692

Find the integrals of the previous equation (problem 5691) for a=0a = 0 as series converging in a neighborhood of the point x=1x = 1.

?
Problem 5693

Find the integral of Bessel's equation

x2y′′+xy′+(x2−n2)y=0, x^{2}y''+xy'+(x^{2}-n^{2})y = 0,

setting y=C1y1+C2y2y = C_{1}y_{1}+C_{2}y_{2}, where y1y_{1} and y2y_{2} are generalized power series of the form xσ(a0+a1x+a2x2+⋯ )x^{\sigma }(a_{0}+a_{1}x+a_{2}x^{2}+\cdots ), where σ\sigma is some constant, not necessarily an integer. The number nn is assumed to be a non-integer.

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Problem 5694

Show that the general integral of Gauss's equation

x(x−1)y′′+((α+β+1)x−γ)y′+αβy=0 x(x-1)y''+\left((\alpha +\beta +1)x-\gamma \right)y'+\alpha \beta y = 0

can, for ∣x∣<1\left|x\right| < 1, be represented by the formula

y=C1F(α,β,γ;x)+C2x1−γF(α−γ+1,β−γ+1,2−γ;x), y = C_{1}F(\alpha ,\beta ,\gamma ;x)+C_{2}x^{1-\gamma }F(\alpha -\gamma +1,\beta -\gamma +1,2-\gamma ;x),

where F(α,β,γ;x)F(\alpha ,\beta ,\gamma ;x) denotes the hypergeometric series:

F(α,β,γ;x)=1+α1⋅βγx+α(α+1)1⋅2⋅β(β+1)γ(γ+1)x2+…. F(\alpha ,\beta ,\gamma ;x) = 1+\frac{\alpha }{1}\cdot \frac{\beta }{\gamma }x+\frac{\alpha (\alpha +1)}{1\cdot 2}\cdot \frac{\beta (\beta +1)}{\gamma (\gamma +1)}x^{2}+\ldots .
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Problem 5695

By means of the Laplace transform, integrate the equation xy′′+(2n+1)y′+xy=0xy''+(2n+1)y'+xy = 0.

?
Problem 5696

By means of the Laplace transform (problem 5695), integrate the equation xy′′−(α+β)(1+x)y′+αβxy=0xy''-(\alpha +\beta )(1+x)y'+\alpha \beta xy = 0.

?
Problem 5697

By means of the Laplace transform (problem 5695), integrate the equation xy′′+2ay′−q2xy=0xy''+2ay'-q^{2}xy = 0; a>0a > 0.

?
Problem 5698

Prove that for 0<a<20 < a < 2 the general integral of the equation xy′′+ay′−q2xy=0xy''+ay'-q^{2}xy = 0 can be represented by the formula

y=∫0πexqcos⁡φ(Asin⁡a−1φ+Bx1−asin⁡1−aφ) dφ, y = \int _{0}^{\pi } e^{xq\cos \varphi }(A\sin ^{a-1}\varphi +Bx^{1-a}\sin ^{1-a}\varphi )\, d\varphi ,

and for a=1a = 1 by the formula:

y=∫0πexqcos⁡φ(A+Bln⁡(xsin⁡2φ)) dφ. y = \int _{0}^{\pi } e^{xq\cos \varphi }\left(A+B\ln (x\sin ^{2}\varphi )\right)\, d\varphi .
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Problem 5699

Find two particular integrals of the equation xy′′′−y=0xy'''-y = 0, setting:

y1=∫sin⁡xt⋅u(t) dt;y2=∫cos⁡xt⋅v(t) dt. y_{1} = \int \sin \frac{x}{t}\cdot u(t)\, dt; \quad y_{2} = \int \cos \frac{x}{t}\cdot v(t)\, dt.
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Problem 5700

Find two pairs of particular integrals of the system

2xy′+2z′+y=0;2y′=2xz′+z;Im(x)<1, 2xy'+2z'+y = 0; \quad 2y' = 2xz'+z; \quad \mathrm{Im}\left(x\right) < 1,

seeking one of the unknowns in the form of the integral ∫u(t)sin⁡xt dt\int u(t)\sin xt\, dt, and the other in the form of the integral ∫v(t)cos⁡xt dt\int v(t)\cos xt\, dt.

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Problem 5701

Prove that the integral of Legendre's equation

(1−x2)y′′−2xy′+n(n+1)y=0 (1-x^{2})y''-2xy'+n(n+1)y = 0

can be represented in one of the forms:

y=∫01t−n+12(1−t)−n+22(x2−t)n2 dt;n>0, y = \int _{0}^{1} t^{-\frac{n+1}{2}}(1-t)^{-\frac{n+2}{2}}(x^{2}-t)^{\frac{n}{2}}\, dt; \quad n > 0, y=∫01tn2(1−t)n−12(x2−t)−n+12 dt;n>−1. y = \int _{0}^{1} t^{\frac{n}{2}}(1-t)^{\frac{n-1}{2}}(x^{2}-t)^{-\frac{n+1}{2}}\, dt; \quad n > -1.
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Problem 5702

Let φ(x,t)=tβ−1(1−t)γ−β−1(1−tx)−α\varphi (x,t) = t^{\beta -1}(1-t)^{\gamma -\beta -1}(1-tx)^{-\alpha }. Prove that each of the integrals

∫1∞φ(x,t) dt;α+1>γ>β; \int _{1}^{\infty } \varphi (x,t)\, dt; \qquad \alpha +1 > \gamma > \beta ; ∫01/xφ(x,t) dt;β>0, α<1; \int _{0}^{1/x} \varphi (x,t)\, dt; \qquad \beta > 0, \ \alpha < 1; ∫01φ(x,t) dt;γ>β>0, α<1; \int _{0}^{1} \varphi (x,t)\, dt; \qquad \gamma > \beta > 0, \ \alpha < 1; ∫0−∞φ(x,t) dt;β>0, α+1>γ, α<1; \int _{0}^{-\infty } \varphi (x,t)\, dt; \qquad \beta > 0, \ \alpha +1 > \gamma , \ \alpha < 1; ∫11/xφ(x,t) dt;γ>β, α<1; \int _{1}^{1/x} \varphi (x,t)\, dt; \qquad \gamma > \beta , \ \alpha < 1; ∫1/x∞φ(x,t) dt;α+1>γ, α<1 \int _{1/x}^{\infty } \varphi (x,t)\, dt; \qquad \alpha +1 > \gamma , \ \alpha < 1

is a particular integral of the hypergeometric equation

x(x−1)y′′+((α+β+1)x−γ)y′+αβy=0. x(x-1)y''+\left((\alpha +\beta +1)x-\gamma \right)y'+\alpha \beta y = 0.
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Problem 5703

Show that the equation

(Ax2+Bx+C)y′′+(Dx+E)y′+Fy=0 (Ax^{2}+Bx+C)y''+(Dx+E)y'+Fy = 0

is reduced, by a simple change of variables, to the hypergeometric equation.

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Problem 5704

Prove the identity: (1+x)n=F(−n,β,β,−x)(1+x)^{n} = F(-n,\beta ,\beta ,-x), where FF is the hypergeometric series (problem 5694).

?
Problem 5705

Prove the identity: ex=lim⁡n→∞F(1,n,1,xn)e^{x} = \lim_{n\to \infty } F\left(1,n,1,\dfrac {x}{n}\right), where FF is the hypergeometric series (problem 5694).

?
Problem 5706

Prove the identity: ln⁡(1+x)=xF(1,1,2,−x)\ln (1+x) = xF(1,1,2,-x), where FF is the hypergeometric series (problem 5694).

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Problem 5707

Prove the identity: 12xln⁡1+x1−x=F(12,1,32,x2)\dfrac {1}{2x}\ln \dfrac {1+x}{1-x} = F\left(\dfrac {1}{2},1,\dfrac {3}{2},x^{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5708

Prove the identity: (1+x)n+(1−x)n=2F(−n2,1−n2,12,x2)(1+x)^{n}+(1-x)^{n} = 2F\left(-\dfrac {n}{2},\dfrac {1-n}{2},\dfrac {1}{2},x^{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5709

Prove the identity: ch⁡x=lim⁡α,β→∞F(α,β,12,x24αβ)\ch x = \lim_{\alpha ,\beta \to \infty } F\left(\alpha ,\beta ,\dfrac {1}{2},\dfrac {x^{2}}{4\alpha \beta }\right), where FF is the hypergeometric series (problem 5694).

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Problem 5710

Prove the identity: cos⁡narccos⁡x=F(n,−n,12,1−x2)\cos n\arccos x = F\left(n,-n,\dfrac {1}{2},\dfrac {1-x}{2}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5711

Prove the identity:

Jn(x)=lim⁡α,β→∞(x2)n1Γ(n+1)F(α,β,n+1,−(x2αβ)2), J_{n}(x) = \lim _{\alpha ,\beta \to \infty } \left(\frac{x}{2}\right)^{n}\frac{1}{\Gamma (n+1)}F\left(\alpha ,\beta ,n+1,-\left(\frac{x}{2\sqrt{\alpha \beta }}\right)^{2}\right),

where FF is the hypergeometric series (problem 5694), and Jn(x)J_{n}(x) is the Bessel function, equal to

∑ν=0∞(−1)ν(x2)n+2νν!Γ(n+ν+1). \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }\left(\dfrac {x}{2}\right)^{n+2\nu }}{\nu !\Gamma (n+\nu +1)}.
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Problem 5712

Denoting F(α+n,β+n,γ+n,1−x2)F\left(\alpha +n,\beta +n,\gamma +n,\dfrac {1-x}{2}\right) (problem 5694) by FnF_{n}, prove that

((1−x)γ(1+x)α+β−γ+1F1)′=−2γ(1−x)γ−1(1+x)α+β−γF0. \left((1-x)^{\gamma }(1+x)^{\alpha +\beta -\gamma +1}F_{1}\right)' = -2\gamma (1-x)^{\gamma -1}(1+x)^{\alpha +\beta -\gamma }F_{0}.
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Problem 5713

Using the preceding (problem 5712), show that

((1−x)γ+n−1(1+x)α+β−γ+nFn)(n)=(−1)n2nγ(γ+1)⋯(γ+n−1)(1−x)γ−1(1+x)α+β−γF0. \left((1-x)^{\gamma +n-1}(1+x)^{\alpha +\beta -\gamma +n}F_{n}\right)^{(n)} = (-1)^{n}2^{n}\gamma (\gamma +1)\cdots (\gamma +n-1)(1-x)^{\gamma -1}(1+x)^{\alpha +\beta -\gamma }F_{0}.
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Problem 5714

Setting γ=β+1\gamma = \beta +1, β=−n\beta = -n, α=p+q+n+1\alpha = p+q+n+1 for the Jacobi polynomial (problem 5581)

Q=n!(p+1)⋯(p+n)Pn(p,q)(x)=F(p+q+n+1,−n,p+1,1−x2), Q = \frac{n!}{(p+1)\cdots (p+n)}P_{n}^{(p,q)}(x) = F\left(p+q+n+1,-n,p+1,\frac{1-x}{2}\right),

prove the identity:

(1−x)p(1+x)qQ=(−1)n2n(p+1)⋯(p+n)dndxn((1−x)p+n(1+x)q+n). (1-x)^{p}(1+x)^{q}Q = \frac{(-1)^{n}}{2^{n}(p+1)\cdots (p+n)}\frac{d^{n}}{dx^{n}}\left((1-x)^{p+n}(1+x)^{q+n}\right).
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Problem 5715

In the particular case p=q=0p = q = 0 (problem 5714), obtain the Rodrigues formula for the Legendre polynomials (problem 5582):

Pn(x)=(−1)n2n⋅n!dn(1−x2)ndxn. P_{n}(x) = \frac{(-1)^{n}}{2^{n}\cdot n!}\frac{d^{n}(1-x^{2})^{n}}{dx^{n}}.
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Problem 5716

Prove the identity for the Chebyshev polynomials (problem 5583):

Tn(x)=cos⁡narccos⁡x2n−1=12n−1F(n,−n,12,1−x2)=(−1)n1⋅3⋯(2n−1)⋅1−x22n−1⋅dn(1−x)n−12dxn. T_{n}(x) = \frac{\cos n\arccos x}{2^{n-1}} = \frac{1}{2^{n-1}}F\left(n,-n,\frac{1}{2},\frac{1-x}{2}\right) = \frac{(-1)^{n}}{1\cdot 3\cdots (2n-1)}\cdot \frac{\sqrt{1-x^{2}}}{2^{n-1}}\cdot \frac{d^{n}(1-x)^{n-\frac{1}{2}}}{dx^{n}}.
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Problem 5717

Prove the identity: ∫0xup−1(1−u)q−1 du=xppF(1−q,p,p+1,x)\displaystyle \int_{0}^{x} u^{p-1}(1-u)^{q-1}\, du = \frac{x^{p}}{p}F(1-q,p,p+1,x), where FF is the hypergeometric series (problem 5694).

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Problem 5718

Prove the identity: ∫0xup−1e−u du=xpplim⁡n→∞F(1−n,p,p+1,xn)\displaystyle \int_{0}^{x} u^{p-1}e^{-u}\, du = \frac{x^{p}}{p}\lim_{n\to \infty } F\left(1-n,p,p+1,\frac{x}{n}\right), where FF is the hypergeometric series (problem 5694).

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Problem 5719

Prove the identity: ∫01du(1−u2)(1−k2u2)=π2F(12,12,1,k2)\displaystyle \int_{0}^{1} \frac{du}{\sqrt{(1-u^{2})(1-k^{2}u^{2})}} = \frac{\pi }{2}F\left(\frac{1}{2},\frac{1}{2},1,k^{2}\right), where FF is the hypergeometric series (problem 5694).

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Section 14.11Applications to Problems of Mathematical Physics[77/69%]
Problem 5720

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=azw = az; a>0a > 0.

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Remark.
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A plane, irrotational motion of an incompressible fluid is characterized by an analytic function w(z)=φ(x,y)+iψ(x,y)w(z) = \varphi (x,y)+i\psi (x,y), where φ\varphi is the velocity potential, and ψ\psi is the stream function. The projections of velocity onto the coordinate axes, vxv_{x} and vyv_{y}, equal the partial derivatives ∂φ∂x\dfrac {\partial \varphi }{\partial x} and ∂φ∂y\dfrac {\partial \varphi }{\partial y}. Hence dwdz=vx−ivy\dfrac {dw}{dz} = v_{x}-iv_{y}. Every fluid particle moves along a streamline ψ(x,y)=C\psi (x,y) = C. The difference of the values of ψ\psi at two points represents the volume of fluid flowing per unit time through a cylinder of unit height whose walls lie on the line joining the given points in the plane XOYXOY. Any streamline, or part of one, may be replaced by a solid wall without disturbing the fluid motion.

Problem 5721

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=iazw = iaz; a>0a > 0.

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Problem 5722

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=z2w = z^{2}.

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Problem 5723

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=1zw = \dfrac {1}{z}.

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Problem 5724

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=1z2w = \dfrac {1}{z^{2}}.

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Problem 5725

Find the lines of equal potential, the streamlines, and the magnitude of the velocity and its components on the axes, if the fluid motion is determined by the motion function w=ln⁡zw = \ln z.

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Problem 5726

The velocity potential has the form

φ=ln⁡∣z2−a2∣−ln⁡∣z2+a2∣. \varphi = \ln \left|z^{2}-a^{2}\right|-\ln \left|z^{2}+a^{2}\right|.

Find the equation of the streamlines, and show that the arcs of the circle ∣z∣=a\left|z\right| = a and the coordinate axes outside the points z=±az = \pm a, z=±iaz = \pm ia are streamlines.

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Problem 5727

The stream function has the form

ψ=ln⁡(r4−2a2r2cos⁡2θ+a4), \psi = \ln (r^{4}-2a^{2}r^{2}\cos 2\theta +a^{4}),

where z=reθiz = re^{\theta i}. Find the potential φ\varphi.

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Problem 5728

The stream function is determined by the equation:

th⁡πxa⋅tg⁡ψ=tg⁡πya;a>0. \th \frac{\pi x}{a}\cdot \tg \psi = \tg \frac{\pi y}{a}; \quad a > 0.

Find the motion function ww.

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Problem 5729

The streamlines are the circles x2+y2=2axx^{2}+y^{2} = 2ax. Find the ratio of the fluid velocity at the points (2a,0)(2a,0) and (a,a)(a,a).

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Problem 5730

The same question (problem 5729, the ratio of the fluid velocity at two points) for the lemniscate r2=a2cos⁡2θr^{2} = a^{2}\cos 2\theta and the points (a,0)(a,0) and (a38,a8)\left(a\sqrt{\dfrac {3}{8}}, \dfrac {a}{\sqrt{8}}\right).

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Problem 5731

For a source z=az = a of strength EE, find the streamlines and the lines of equal potential.

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Remark.
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Call the flux through a closed contour CC the quantity of fluid EE flowing per unit time through a cylinder of unit height whose base lies along the contour CC. The quantity EE is given by the formula:

E=∫Cvx dy−vy dx. E = \int _{C} v_{x}\, dy-v_{y}\, dx.

The circulation of the velocity along the same contour CC is called the quantity Γ=∫Cvx dx+vy dy\Gamma = \int_{C} v_{x}\, dx+v_{y}\, dy. Since Γ+iE=∫Cdwdz⋅dz\Gamma +iE = \int_{C} \dfrac {dw}{dz}\cdot dz, EE and Γ\Gamma are conveniently found by means of residues. If w=E2πln⁡(z−a)w = \dfrac {E}{2\pi }\ln (z-a), where EE is real, then the point aa is called a source of strength EE. If w=Γ2πiln⁡(z−a)w = \dfrac {\Gamma }{2\pi i}\ln (z-a), where Γ\Gamma is real, then the point aa is called a vortex of intensity Γ\Gamma.

Problem 5732

The fluid motion is produced by a source of strength EE (problem 5731), placed at the point aa. Prove that the flux through a simple closed contour CC equals zero if the point aa is outside the contour, and equals EE if the point aa is inside the contour.

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Problem 5733

Under the same conditions (problem 5732), prove that the circulation along any closed contour equals zero.

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Problem 5734

Obtain the corresponding results (problems 5732--5733) for the motion produced by a vortex of intensity Γ\Gamma at the point aa (problem 5731).

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Problem 5735

The fluid motion is produced by a source EE and a vortex Γ\Gamma, placed at the point aa (problem 5731). Prove that the streamlines are logarithmic spirals.

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Problem 5736

The motion function is w=ln⁡sh⁡πzw = \ln \sh \pi z. Find the flux through the circle 2∣z∣=32\left|z\right| = 3 and the circulation along it.

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Problem 5737

The motion function is w=2iln⁡(z2−a2)w = 2i\ln (z^{2}-a^{2}). Find the circulation along the circles ∣z−a∣=a\left|z-a\right| = a; ∣z+a∣=a\left|z+a\right| = a.

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Problem 5738

The motion function is w=ln⁡(z2+z−2)w = \ln (z^{2}+z^{-2}). What are the sources?

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Problem 5739

Prove that the real axis is a streamline, if the motion is created by sources of strength EE, placed at the points a±bia \pm bi.

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Problem 5740

Prove that the reflection in the real axis of a vortex Γ\Gamma at the point a+bia+bi is a vortex −Γ-\Gamma at the point a−bia-bi.

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Remark.
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Remark. If a streamline CC divides the plane into two parts, and the motion is created by sources and vortices, then the collection of sources and vortices on one side of the line CC is called the reflection of the sources and vortices on the other side of the same line.

Problem 5741

At the point aa outside the disk ∣z∣=R\left|z\right| = R there is a source EE. Prove that its reflection in the circle (problem 5740) consists of two sources: a source EE at the point R2a‾\dfrac {R^{2}}{\overline{a}} and a source of strength −E-E at the point z=0z = 0.

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Problem 5742

A vortex of intensity Γ\Gamma is placed outside the disk ∣z∣=R\left|z\right| = R. Prove that its reflection in the circle (problem 5740) is a vortex −Γ-\Gamma at the point R2a‾\dfrac {R^{2}}{\overline{a}}.

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Problem 5743

Find the fluid motion outside the disk ∣z∣=1\left|z\right| = 1, produced by a source of strength EE at the point aa, where a>1a > 1.

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Problem 5744

The same (problem 5743) for a vortex of intensity Γ\Gamma.

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Problem 5745

Find the fluid motion in the domain ∣z∣>1\left|z\right| > 1, Im(z)>0\mathrm{Im}\left(z\right) > 0, produced by a vortex at the point z=aiz = ai, a>1a > 1, of intensity Γ\Gamma.

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Problem 5746

The positive parts of the coordinate axes are impermeable walls. Only at the point z=1z = 1 is there a point source (more precisely, a source along a slit perpendicular to the plane XOYXOY and passing through z=1z = 1), from which, in a space of height 11, QQ units flow into the first quadrant per second. Find the fluid motion.

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Problem 5747

Find the fluid motion in the region 0<y<x30 < y < x\sqrt{3}, produced by a source of strength EE at the point z=aeπi6z = ae^{\frac{\pi i}{6}}; a>0a > 0.

?
Problem 5748

Find the fluid motion in the first quadrant, if at the points z=1z = 1 and z=iz = i in the walls of the quadrant there are point sources through which, respectively, QQ units of volume of fluid flow in and out per second.

?
Problem 5749

Sources of strength ±Ea\pm \dfrac {E}{a} are located at the points z=±az = \pm a, where a→0a \to 0 (a dipole). Find the limiting fluid motion.

?
Problem 5750

On a circle, impermeable to fluid elsewhere, two sources of strength EE and −E-E are placed. Prove that the streamlines inside the disk are arcs of circles passing through both sources.

?
Problem 5751

Two vortices Γ\Gamma and −Γ-\Gamma are placed at two points. Prove that in the motion produced by them, the streamlines are circles.

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Problem 5752

The fluid motion is produced by a stream from infinity, with velocity equal to aa and directed in the negative direction of the axis OxOx, and by a source E=1E = 1 at the origin. Prove that the fluid issuing from the source fills the region inside the curve xtg⁡2πay=yx\tg 2\pi ay = y.

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Problem 5753

The motion in the half-plane y>0y > 0 is produced by a source at the point z=aiz = ai. Prove that the streamlines are the hyperbolas x2−y2+a2=Cxyx^{2}-y^{2}+a^{2} = Cxy.

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Problem 5754

The fluid motion is created by the superposition of sources at z=0,±πi,±2πi,…z = 0, \pm \pi i, \pm 2\pi i, \ldots with strength E=2πE = 2\pi each. Prove that the velocity of motion tends to unity as one recedes from OYOY, and is directed away from the axis OYOY almost perpendicularly to it.

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Problem 5755

At the same points (problem 5754) there are vortices of intensity Γ\Gamma. Prove that at points far from the axis OYOY the velocity in magnitude is close to Γ2π\dfrac {\Gamma }{2\pi }, and in direction is almost parallel to the axis OYOY.

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Problem 5756

The fluid motion is created by the superposition of sources of strength E=2πE = 2\pi at the points z=0,±πi,±2πi,…z = 0, \pm \pi i, \pm 2\pi i, \ldots and a stream of velocity v>1v > 1, parallel to the negative direction of the axis OXOX. Prove that the greatest abscissa of the points reached by the fluid from the sources equals 12ln⁡v+1v−1\dfrac {1}{2}\ln \dfrac {v+1}{v-1}.

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Problem 5757

Prove that, under the conditions of the previous problem (problem 5756), the region filled by the fluid from the source z=0z = 0 is bounded by the curve cth⁡x⋅tg⁡y=tg⁡vy\cth x\cdot \tg y = \tg vy, the distance between whose branches tends to 2πv+1\dfrac {2\pi }{v+1} as x→−∞x \to -\infty.

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Problem 5758

A source of strength E=2πE = 2\pi is located at the point z=aiz = ai, where 0<a<π20 < a < \dfrac {\pi }{2}. Prove that the motion it creates in the strip 0<y<π20 < y < \dfrac {\pi }{2} has motion function:

w=ln⁡sh⁡((z+ia)sh⁡(z−ia)). w = \ln \sh \left((z+ia)\sh (z-ia)\right).
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Problem 5759

A source of strength E=2πE = 2\pi is located at the point z0=a+biz_{0} = a+bi, where a>0a > 0; 0<b<π20 < b < \dfrac {\pi }{2}. Prove that the motion function it creates in the domain 0<y<π20 < y < \dfrac {\pi }{2}; x>0x > 0, is given by the formula:

w=ln⁡(sh⁡(z−z0)sh⁡(z+z0)sh⁡(z−z0‾)sh⁡(z+z0‾)). w = \ln \left(\sh (z-z_{0})\sh (z+z_{0})\sh (z-\overline{z_{0}})\sh (z+\overline{z_{0}})\right).
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Problem 5760

The fluid motion in the region ∣z∣>1\left|z\right| > 1, produced by a source of strength EE at the point a>1a > 1, is characterized by the function w=E2πln⁡(z+1z−a−1a)w = \dfrac {E}{2\pi }\ln \left(z+\dfrac {1}{z}-a-\dfrac {1}{a}\right). Setting E=2πb(a+1a)E = 2\pi b\left(a+\dfrac {1}{a}\right) and a→∞a \to \infty, find the motion arising from flow past the disk ∣z∣<1\left|z\right| < 1 by a stream whose velocity at infinity is v∞=vv_{\infty } = v and is directed toward the negative axis OXOX.

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Problem 5761

Along the real axis there is an impermeable wall. At a distance hh from it there is a vortex of intensity Γ\Gamma. Prove that it moves parallel to the axis OXOX with velocity Γ4πh\dfrac {\Gamma }{4\pi h}.

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Remark.
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If the fluid motion in a simply connected domain, determined by a function w(z)w(z), is produced by a vortex of intensity Γ\Gamma, located at the given moment at the point z=az = a, then this vortex is displaced with velocity

dadt=(dw1dz‾)z=a, \frac{da}{dt} = \left(\overline{\frac{dw_{1}}{dz}}\right)_{z=a},

where

w1=w−Γ2πiln⁡(z−a)=φ(x,y)+iψ(x,y). w_{1} = w-\frac{\Gamma }{2\pi i}\ln (z-a) = \varphi (x,y)+i\psi (x,y).

The trajectory of the vortex is given by the equation

ψ(ξ,η)=C,where a=ξ+iη, \psi (\xi ,\eta ) = C, \quad \text{where } a = \xi +i\eta ,

and the fluid velocity at infinity is assumed to be zero.

Problem 5762

Prove that a vortex, placed inside an impermeable circle, moves along a concentric circle.

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Problem 5763

The motion in the region x>0x > 0, y>0y > 0 is produced by a vortex of intensity Γ\Gamma, placed at the point (ξ,η)(\xi ,\eta ). Prove that the trajectory of the vortex is expressed by the equation ξ−2+η−2=C\xi^{-2}+\eta^{-2} = C, and the components of the velocity of the vortex are given by the formulas:

dξdt=Γξ24πη(ξ2+η2);dηdt=−Γη24πξ(ξ2+η2). \frac{d\xi }{dt} = \frac{\Gamma \xi ^{2}}{4\pi \eta (\xi ^{2}+\eta ^{2})}; \quad \frac{d\eta }{dt} = -\frac{\Gamma \eta ^{2}}{4\pi \xi (\xi ^{2}+\eta ^{2})}.
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Problem 5764

Prove that the motion function produced by an infinite chain of vortices, each of intensity Γ=2π\Gamma = 2\pi, placed at the points z=0,±π,±2π,…z = 0, \pm \pi , \pm 2\pi , \ldots, is expressed by the formula

w(z)=−iln⁡sin⁡z. w(z) = -i\ln \sin z.
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Problem 5765

Find the motion function produced by the presence of vortices of two kinds: a) with intensity Γ\Gamma at the points l(n+14)+hi2l\left(n+\dfrac {1}{4}\right)+\dfrac {hi}{2}, b) with intensity −Γ-\Gamma at the points l(n−14)−hi2l\left(n-\dfrac {1}{4}\right)-\dfrac {hi}{2}. Here n=0,1,2,3,…n = 0,1,2,3,\ldots.

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Problem 5766

Find the velocity of a vortex of intensity Γ\Gamma, placed between parallel walls, at a distance hh from one of them, if the distance between the walls is dd.

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Problem 5767

Find the motion in the region −π2<x<π2-\dfrac {\pi }{2} < x < \dfrac {\pi }{2}, if at the point (0,0)(0,0) there is a source of strength QQ.

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Remark.
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If in the identity w(z)=u(x,y)+iv(x,y)w(z) = u(x,y)+iv(x,y); z=x+iyz = x+iy, one introduces a new variable ζ=ξ+iη\zeta = \xi +i\eta, setting

z=φ(ζ)=p(ξ,η)+iq(ξ,η), z = \varphi (\zeta ) = p(\xi ,\eta )+iq(\xi ,\eta ),

then, denoting the quantities w(z)w(z), u(x,y)u(x,y), v(x,y)v(x,y), expressed in terms of ζ\zeta, ξ\xi, and η\eta, by w1(ζ)w_{1}(\zeta ), u1(ξ,η)u_{1}(\xi ,\eta ), v1(ξ,η)v_{1}(\xi ,\eta ), we obtain w1(ζ)=u1(ξ,η)+iv1(ξ,η)w_{1}(\zeta ) = u_{1}(\xi ,\eta )+iv_{1}(\xi ,\eta ). After this, the domain SS in the plane zz goes into a domain S1S_{1} in the plane ζ\zeta, and the streamlines v(x,y)=Cv(x,y) = C go into the streamlines v1(ξ,η)=Cv_{1}(\xi ,\eta ) = C.

Thus, a conformal mapping makes it possible, from the motion in the domain SS, given by the function w(z)w(z), to find the motion in the domain S1S_{1} by means of the function w1(ζ)w_{1}(\zeta ).

Problem 5768

Find the motion in the region −π2<x<π2-\dfrac {\pi }{2} < x < \dfrac {\pi }{2} (problem 5767), if the source is located at the point (π2,h)\left(\dfrac {\pi }{2},h\right).

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Problem 5769

Find the motion function for symmetric, continuous flow past a cylinder with circular base ∣z∣=a\left|z\right| = a, if the velocity of the stream at infinity equals vv and is directed toward the negative axis OXOX. (One method of solution: map the region ∣z∣>a\left|z\right| > a, Im(z)>0\mathrm{Im}\left(z\right) > 0 onto the upper half-plane of a new variable.)

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Problem 5770

Find the motion function for flow past the segment between the points z=±iz = \pm i by the same stream as before (problem 5769). The flow sought is continuous and symmetric.

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Problem 5771

The function w(z)=a(z+R2z)+Γ2πiln⁡zw(z) = a\left(z+\dfrac {R^{2}}{z}\right)+\dfrac {\Gamma }{2\pi i}\ln z gives an asymmetric, continuous flow past the disk ∣z∣=R\left|z\right| = R. Find v∞v_{\infty }, the circulation along the circle, and the points where the velocity is zero.

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Problem 5772

Find the motion function for the flow of fluid from the half-plane y>0y > 0 into the half-plane y<0y < 0 through an opening in the wall y=0y = 0 between the points z=±1z = \pm 1. The velocity at infinity equals zero.

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Problem 5773

Find the stream function w(z)w(z) for continuous, symmetric flow past an elliptic cylinder with cross-section b2x2+a2y2=a2b2b^{2}x^{2}+a^{2}y^{2} = a^{2}b^{2} by a stream having, at infinity, velocity vx=vv_{x} = v; vy=0v_{y} = 0.

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Problem 5774

Prove that the function w=iln⁡(1+z−1)w = i\ln (1+z^{-1}) gives the flow of fluid past a region bounded by a line and a circle, under the condition v∞=0v_{\infty } = 0.

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Problem 5775

Find the function giving flow past the half-disk ∣z+43∣<23\left|z+\dfrac {4}{3}\right| < \dfrac {2}{3}, Im(z)>0\mathrm{Im}\left(z\right) > 0, under the condition at infinity vx=−vv_{x} = -v, vy=0v_{y} = 0.

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Problem 5776

Show that the function w=2iδln⁡Rzw = 2i\delta \ln \dfrac {R}{z} determines the electric field outside a circular cylinder ∣z∣=R\left|z\right| = R, on which a charge equal to δ\delta is uniformly distributed for a cylinder of unit height.

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Remark.
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A plane electrostatic field, in the absence of continuously distributed charges, is characterized by an analytic function w(z)=u(x,y)+iv(x,y)w(z) = u(x,y)+iv(x,y). Here

idwdz=Ex−iEy, i\frac{dw}{dz} = E_{x}-iE_{y},

where ExE_{x} and EyE_{y} are the projections of the electric force onto the coordinate axes. The equation u(x,y)=Cu(x,y) = C gives the lines of force, and v(x,y)=Cv(x,y) = C gives the equipotential lines. Inside and on the boundary of a conductor the potential vv must be constant. The density of electricity at any point of the contour of a conductor is given, up to sign, by the formula: σ=14π∣dwdz∣\sigma = \dfrac {1}{4\pi }\left|\dfrac {dw}{dz}\right|.

Problem 5777

Find the lines of force and the equipotential lines in the previous problem (problem 5776).

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Problem 5778

Find the force field between the surfaces of two cylinders with circular bases ∣z∣=1\left|z\right| = 1, ∣z−1∣=4\left|z-1\right| = 4, if the potential on the first equals 11, and on the second, zero.

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Problem 5779

On an isolated cylinder with circular cross-section ∣z−2i∣=1\left|z-2i\right| = 1, the density of electricity is σ=1\sigma = 1. How does it distribute if a plane, parallel to the generator and intersecting the plane XOYXOY along the real axis, is grounded (electrostatic induction)?

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Problem 5780

On the cylinder ∣z−5∣=4\left|z-5\right| = 4 the potential equals 11, and on the cylinder ∣z+5∣=4\left|z+5\right| = 4 it equals zero. Find the greatest and least densities of electricity on both cylinders.

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Problem 5781

At the point z=2z = 2 there is a point charge δ\delta. The imaginary axis is grounded. What is the potential at the point z=1z = 1?

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Problem 5782

Representing the equation of the parabola y2=2pxy^{2} = 2px in parametric form:

x=2pu2,y=2pu, x = 2pu^{2}, \qquad y = 2pu,

prove that the function ww, defined by the identity z=2p(w2+iw)z = 2p(w^{2}+iw), gives the field outside the parabola, in which the potential on the parabola itself equals zero and in which there are no charges.

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Problem 5783

Find the field outside the ellipse

x=acos⁡u,y=bsin⁡u, x = a\cos u, \qquad y = b\sin u,

representing one of the equipotential lines of the field.

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Problem 5784

The potential equals ν0\nu_{0} at the points of the line z=x+ihz = x+ih and equals zero on the real axis. Find the function w(z)w(z) giving the electrostatic field.

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Problem 5785

On the line 3x2+4y2=123x^{2}+4y^{2} = 12 the potential equals one, and on the segment between the foci it equals zero. Find the potential at the point z=iz = i.

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Problem 5786

Find the greatest and least densities on the ellipse of the previous problem (problem 5785).

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Problem 5787

On the line z=x>1z = x > 1 the potential equals ν0\nu_{0}, and on the line z=x<−1z = x < -1 it equals −ν0-\nu_{0}. Find the function w(z)w(z) giving the electrostatic field.

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Problem 5788

On the line z=iyz = iy, where y>1y > 1, the potential equals 11, and on the real axis it equals zero. Find the density distribution of electricity on the axis OXOX.

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Problem 5789

The potential equals π\pi on the polygonal line consisting of the lines:

z=x+πi;x<0andz=iy;y>π. z = x+\pi i; \quad x < 0 \quad \text{and} \quad z = iy; \quad y > \pi .

On the axis OXOX the potential equals zero. Find the electrostatic field in the region bounded by the indicated lines, using the Christoffel--Schwarz integral.

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Remark.
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If on a contour CC there is a charge qq, and if the region outside the contour is mapped conformally onto the region outside the disk ∣ζ∣=1\left|\zeta \right| = 1 by an analytic function

z=a1ζ+a0+∑n=1∞a−nζ−n, z = a_{1}\zeta +a_{0}+\sum _{n=1}^{\infty } a_{-n}\zeta ^{-n},

then the field outside CC is given by the function

w=2iqln⁡w0ζ(z), w = 2iq\ln \frac{w_{0}}{\zeta (z)},

where w0w_{0} is some constant.

Problem 5790

Find the field outside the ellipse b2x2+a2y2=a2b2b^{2}x^{2}+a^{2}y^{2} = a^{2}b^{2}.

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Problem 5791

Find the field outside the square −1<x<1-1 < x < 1, −1<y<1-1 < y < 1, on which a charge qq is placed.

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Problem 5792

On the circle ∣z∣=1\left|z\right| = 1 the temperature is 0°0°, and on the circle ∣z−1∣=52\left|z-1\right| = \dfrac {5}{2} it is 100°100°. Find the temperature tt in the annulus between the circles.

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Remark.
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The plane problem of the steady-state distribution of heat inside a homogeneous body is given by an analytic function

w(z)=u(x,y)+iv(x,y), w(z) = u(x,y)+iv(x,y),

where v(x,y)v(x,y) is the temperature at the point (x,y)(x,y). The equation v(x,y)=Cv(x,y) = C gives the isothermal lines, and u(x,y)=Cu(x,y) = C the heat-flow lines.

Problem 5793

On the circle ∣z−5i∣=4\left|z-5i\right| = 4 the temperature is 100°100°, and on the axis OXOX it is zero. Find the temperature tt at the point z=i+1z = i+1.

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Problem 5794

Find the temperature distribution in the region 0<φ<π0 < \varphi < \pi, r>ar > a, where z=reφiz = re^{\varphi i}, if the temperature equals zero on the real axis and one on the half-circle z=aeφiz = ae^{\varphi i}, 0<φ<π0 < \varphi < \pi.

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Problem 5795

Find the heat distribution inside the half-disk ∣z∣<1\left|z\right| < 1, 0<arg⁡z<π0 < \arg z < \pi, if the temperature equals zero on the semicircle and one on the diameter.

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Problem 5796

The same problem (problem 5795), but for a sector of a circle: 0<φ<α0 < \varphi < \alpha, 0<r<a0 < r < a, where z=reφiz = re^{\varphi i}, with the temperature 0°0° on the radii and 1°1° on the arc of the circle.

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