Taylor and Laurent expansions of regular branches of multivalued functions

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Find a Laurent series that converges in the annulus 1<∣z∣<21<\left|z\right|<2 to a branch of the function log⁡(z(2−z)1−z)\log \left(\dfrac {z(2-z)}{1-z}\right).

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Let f(z)f(z) be the ``branch'' of log⁡z\log z defined off the negative real axis so that f(1)=0f(1) = 0.

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(a)

Find the Taylor polynomial of ff of degree 22 at −4+3i-4+3i, simplifying the coefficients.

(b)

Find the radius of convergence RR of the Taylor series T(z)T(z) of f(z)f(z) at −4+3i-4+3i.

(c)

Identify on a picture any points zz where T(z)T(z) converges but T(z)≠f(z)T(z) \neq f(z), and describe the relationship between ff and TT at such points. If there are no such points, is this something special to this example, or a general impossibility? Explain and/or give examples.

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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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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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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There is a region Ω\Omega that exp⁡(Ω)=D(1;1)\exp (\Omega ) = D(1 ; 1). Show that exp⁡\exp is one-to-one in Ω\Omega, but that there are many such Ω\Omega. Fix one, and define log⁡z\log z, for ∣z−1∣<1\left|z-1\right| < 1, to be that w˙∈Ω\dot{w} \in \Omega for which ew=ze^{w} = z. Prove that log⁡′(z)=1/z\log^{\prime }(z) = 1 / z. Find the coefficients ana_{n} in

1z=∑n=0∞an(z−1)n \frac{1}{z} = \sum _{n = 0}^{\infty } a_{n}(z-1)^{n}

and hence find the coefficients cnc_{n} in the expansion

log⁡z=∑n=0∞cn(z−1)n \log z = \sum _{n = 0}^{\infty } c_{n}(z-1)^{n}

In what other discs can this be done?

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Let φ(z)\varphi (z) be a regular branch of the multivalued function Ln⁡z\operatorname {Ln} z in the domain shown in Fig. 18.4, satisfying the condition φ(1)=0\varphi (1) = 0. Find φ′(2)\varphi^{\prime }(2). Expand the function φ(z)\varphi (z) in a Taylor series in a neighborhood of the point z=−3z = -3 in powers of z+3z+3.

Fig. 18.4Fig. 18.4

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Let g(z)g(z) be a regular branch of the multivalued function {z3}\left\{ \sqrt[3]{z}\right\} in the domain shown in Fig. 18.4, such that g(−1)=−1g(-1) = -1. Find g(−2),g′(−3)g(-2), g^{\prime }(-3). Expand g(z)g(z) in a Taylor series in a neighborhood of the point z=2z = 2 in powers of z−2z-2.

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Let φ1(z),φ2(z)\varphi_{1}(z), \varphi_{2}(z) be the functions from Problem 4. Find φ1′(−2),φ2′(2)\varphi_{1}^{\prime }(-2), \varphi_{2}^{\prime }(2). Expand the function φ2(z)\varphi_{2}(z) in a Taylor series in a neighborhood of the point z=4z = 4 in powers of (z−4)(z-4). Expand φ1(z)\varphi_{1}(z) and φ2(z)\varphi_{2}(z) in Laurent series in a neighborhood of the point z=∞z = \infty.

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Relying on the Taylor series expansion of the regular branches of the multivalued function {(1+z)a}\left\{ (1+z)^{a}\right\} (see Example 2 in the reference material), prove the formulas for all the resulting regular branches g(z),w(z),h(z)g(z), w(z), h(z) in the disk ∣z∣<1\left|z\right| < 1:

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(1)

1g(z)=1g(0)+∑n=1+∞(2n)!22n(n!)2z2n\frac{1}{g(z)} = \frac{1}{g(0)}+\sum_{n = 1}^{+\infty } \frac{(2 n)!}{2^{2 n}(n!)^{2}} z^{2 n}, where g(z)∈{1−z2}g(z) \in \left\{ \sqrt{1-z^{2}}\right\};

(2)

g(z)=g(0)+∑n=1+∞(−1)n+1(2n−2)!22n−1n!(n−1)!z2ng(z) = g(0)+\sum_{n = 1}^{+\infty }(-1)^{n+1} \frac{(2 n-2)!}{2^{2 n-1} n!(n-1)!} z^{2 n}, where g(z)∈{1+z2}g(z) \in \left\{ \sqrt{1+z^{2}}\right\};

(3)

w(z)=w(0)+∑n=0+∞(2n)!22n(n!)2(2n+1)z2n+1w(z) = w(0)+\sum_{n = 0}^{+\infty } \frac{(2 n)!}{2^{2 n}(n!)^{2}(2 n+1)} z^{2 n+1}, where w(z)∈Arcsin⁡zw(z) \in \operatorname {Arcsin} z;

(4)

h(z)=h(0)+∑n=0+∞(−1)n(2n)!z2n+122n(n!)2(2n+1)h(z) = h(0)+\sum_{n = 0}^{+\infty }(-1)^{n} \frac{(2 n)!z^{2 n+1}}{2^{2 n}(n!)^{2}(2 n+1)}, where h(z)∈Ln⁡(z+g(z))h(z) \in \operatorname {Ln}(z+g(z)), g(z)∈{1+z2}, g(0)=1g(z) \in \left\{ \sqrt{1+z^{2}}\right\} , ~ g(0) = 1.

For all regular branches h(z),w(z)h(z), w(z) of the multivalued functions in the disk ∣z∣<1\left|z\right| < 1, prove the formulas:

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(1)

h(z)=h(0)+2∑n=0+∞z2n+12n+1h(z) = h(0)+2 \sum_{n = 0}^{+\infty } \frac{z^{2 n+1}}{2 n+1}, where h(z)∈Ln⁡1+z1−zh(z) \in \operatorname {Ln} \frac{1+z}{1-z};

(2)

w(z)=w(0)+∑n=0+∞(−1)nz2n+12n+1w(z) = w(0)+\sum_{n = 0}^{+\infty }(-1)^{n} \frac{z^{2 n+1}}{2 n+1}, where w(z)∈Arctg⁡zw(z) \in \operatorname {Arctg} z;

(3)

1−zzh(z)=−1+∑n=1+∞znn(n+1)\frac{1-z}{z} h(z) = -1+\sum_{n = 1}^{+\infty } \frac{z^{n}}{n(n+1)}, where h(z)∈Ln⁡(1−z),h(0)=0h(z) \in \operatorname {Ln}(1-z), h(0) = 0.

Find the Taylor series expansion in powers of zz in a neighborhood of the point z=0z = 0 of the following functions, containing regular branches g(z),h(z)g(z), h(z) of multivalued functions:

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(1)

sh⁡g(z)⋅sin⁡g(z)\operatorname {sh} g(z) \cdot \sin g(z), where g(z)∈{z},g(0)=0g(z) \in \left\{ \sqrt{z}\right\} , g(0) = 0;

(2)

12g(z)h(1+g(z)1−g(z))\frac{1}{2 g(z)} h\left(\frac{1+g(z)}{1-g(z)}\right), where g(z)∈{z},g(0)=0,h(z)∈Ln⁡z,h(1)=0g(z) \in \left\{ \sqrt{z}\right\} , g(0) = 0, h(z) \in \operatorname {Ln} z, h(1) = 0;

(3)

1g(z)\frac{1}{g(z)}, where g(z)∈{1−z3},g(0)=1g(z) \in \left\{ \sqrt{1-z^{3}}\right\} , g(0) = 1;

(4)

h(z)\quad h(z), where h(z)∈Ln⁡(1+z+z2),h(0)=0h(z) \in \operatorname {Ln}\left(1+z+z^{2}\right), h(0) = 0.

Using the fact that the regular function f(z)∈{Arcsin⁡z1−z2}f(z) \in \left\{ \frac{\operatorname {Arcsin} z}{\sqrt{1-z^{2}}}\right\} satisfies the differential equation

(1−z2)f′(z)−zf(z)=1,f(0)=0 \left(1-z^{2}\right) f^{\prime }(z)-z f(z) = 1, \quad f(0) = 0

prove the formula

f(z)=∑n=0+∞(−1)n(2n)!!(2n+1)!!z2n+1,∣z∣<1. f(z) = \sum _{n = 0}^{+\infty }(-1)^{n} \frac{(2 n)!!}{(2 n+1)!!} z^{2 n+1}, \quad \left|z\right| < 1.
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For the regular branch

f(z)∈{(Arctg⁡zz)2},f(0)=1, f(z) \in \left\{ \left(\frac{\operatorname {Arctg} z}{z}\right)^{2}\right\} , \quad f(0) = 1,

prove the formula

f(z)=1+∑n=1+∞(−1)n(1+13+…+12n+1)z2nn+1,∣z∣<1. f(z) = 1+\sum _{n = 1}^{+\infty }(-1)^{n}\left(1+\frac{1}{3}+\ldots +\frac{1}{2 n+1}\right) \frac{z^{2 n}}{n+1}, \quad \left|z\right| < 1.
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Using the binomial series, prove the validity of the Laurent series expansion formulas for regular branches f(z)f(z) of multivalued functions:

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(1)

f(z)=1+∑n=−∞−1a(a+1)…(a−n−1)(−n)!znf(z) = 1+\sum_{n = -\infty }^{-1} \frac{a(a+1) \ldots (a-n-1)}{(-n)!} z^{n}, where

f(z)∈{(zz−1)a},f(∞)=1,∣z∣>1 f(z) \in \left\{ \left(\frac{z}{z-1}\right)^{a}\right\} , \quad f(\infty ) = 1, \quad \left|z\right| > 1
(2)

f(z)=1z+∑n=−∞−112⋅32…(12−n−1)(−n)!z2n−1f(z) = \frac{1}{z}+\sum_{n = -\infty }^{-1} \frac{\frac{1}{2} \cdot \frac{3}{2} \ldots \left(\frac{1}{2}-n-1\right)}{(-n)!} z^{2 n-1}, where

f(z)∈{1z2−1},f(2)=13,∣z∣>1 f(z) \in \left\{ \frac{1}{\sqrt{z^{2}-1}}\right\} , \quad f(2) = \frac{1}{\sqrt{3}}, \quad \left|z\right| > 1
(3)

f(z)=1+∑n=−∞−1α(α+1)…(α−n−1)(−n)!(b−a)−n⋅(z−a)nf(z) = 1+\sum_{n = -\infty }^{-1} \frac{\alpha (\alpha +1) \ldots (\alpha -n-1)}{(-n)!}(b-a)^{-n} \cdot (z-a)^{n}, where a≠b,f(z)∈{(z−az−b)α},f(∞)=1,∣z−a∣>∣b−a∣;a \neq b, \quad f(z) \in \left\{ \left(\frac{z-a}{z-b}\right)^{\alpha }\right\} , \quad f(\infty ) = 1, \quad \left|z-a\right| > \left|b-a\right| ;

(4)

f(z)=ln⁡2+∑n=−∞−1(−1)n+12n⋅12⋅32⋯(12−n−1)(−n)!z2nf(z) = \ln 2+\sum_{n = -\infty }^{-1} \frac{(-1)^{n+1}}{2 n} \cdot \frac{\frac{1}{2} \cdot \frac{3}{2} \cdots \left(\frac{1}{2}-n-1\right)}{(-n)!} z^{2 n}, where f(z)∈Ln⁡z+z2+1z,f(2)=ln⁡(1+52),∣z∣>1f(z) \in \operatorname {Ln} \frac{z+\sqrt{z^{2}+1}}{z}, \quad f(2) = \ln \left(1+\frac{\sqrt{5}}{2}\right), \quad \left|z\right| > 1.

Verify that the following multivalued functions admit the extraction of regular branches in the annulus GG, and expand all the regular branches in a Laurent series in powers of z−az-a in the annulus GG (the point aa and the annulus GG are indicated in parentheses):

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(1)

Ln⁡(z−1)(z−2)(z+1)(z+2)(a=0,G={z:1<∣z∣<2})\operatorname {Ln} \frac{(z-1)(z-2)}{(z+1)(z+2)} \quad (a = 0, G = \left\{ z: 1 < \left|z\right| < 2\right\} );

(2)

Ln⁡(z+1)2z2+4(a=0,G={z:∣z∣>2})\operatorname {Ln} \frac{(z+1)^{2}}{z^{2}+4}(a = 0, G = \left\{ z:\left|z\right| > 2\right\} );

(3)

Ln⁡z(z+3)(z+2)(z−1)(a=−1,G={z:1<∣z+1∣<2})\operatorname {Ln} \frac{z(z+3)}{(z+2)(z-1)}(a = -1, G = \left\{ z: 1 < \left|z+1\right| < 2\right\} );

(4)

Ln⁡(z−1)2(z+2)(z+3)(a=−1,G={z:∣z+1∣>2})\operatorname {Ln} \frac{(z-1)^{2}}{(z+2)(z+3)}(a = -1, G = \left\{ z:\left|z+1\right| > 2\right\} ).

Let g(z)g(z) be a regular branch of the multivalued function {z+93}\left\{ \sqrt[3]{z+9}\right\} in the plane with a cut along the curve z=9eit,−π⩽t⩽π2z = 9 e^{i t}, \quad -\pi \leqslant t \leqslant \frac{\pi }{2} and the ray z=9i+tiz = 9 i+t i, t⩾0t \geqslant 0, such that arg⁡g(10)=2π3\arg g(10) = \frac{2 \pi }{3}. Compute g(−8),g(−1),g(−9+8i),g′(0)g(-8), g(-1), g(-9+8 i), g^{\prime }(0). Expand g(z)g(z) in a Taylor series in a neighborhood of the point z=3z = 3 in powers of z−3z-3 and draw the largest domain in which this series converges to the function g(z)g(z).

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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡(2−z)\operatorname {Ln}(2-z) in the plane with a cut along the curve z=2eit,0⩽t⩽32πz = 2 e^{i t}, 0 \leqslant t \leqslant \frac{3}{2} \pi, and the ray z=−2i+t,t⩾0z = -2 i+t, t \geqslant 0, such that Im(h(−3))=0\mathrm{Im}\left(h(-3)\right) = 0. Compute h(−2−0),h(−2+0),h(2+i)h(-2-0), \quad h(-2+0), \quad h(2+i), h′(0)h^{\prime }(0). Expand the function h(z)h(z) in a Taylor series centered at the point z=−1z = -1. Find the radius of convergence of this series. Draw the largest domain in which the series converges to the function h(z)h(z).

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Let g(z)g(z) be a regular branch of the multivalued function {z(2−z)23}\left\{ \sqrt[3]{z(2-z)^{2}}\right\} in the plane with a cut along the segment [0,2][0,2], such that g(1+i0)=1g(1+i 0) = 1. Find g(1−i0),g(−3),g′(3)g(1-i 0), \quad g(-3), \quad g^{\prime }(3). Expand g(z)g(z) in a Laurent series in powers of zz in a neighborhood of the point z=∞z = \infty.

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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡z+13−z\operatorname {Ln} \frac{z+1}{3-z} in the plane with a cut along the segment [−1,3][-1,3], such that h(1+i0)=0h(1+i 0) = 0. Find h(1−i0)h(1-i 0), g(−2),g′(−3)g(-2), g^{\prime }(-3). Expand h(z)h(z) in a Taylor series centered at the point z=5z = 5 and in a Laurent series in powers of zz in a neighborhood of the point z=∞z = \infty. Find the domains of convergence of these series.

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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡z+iz+1\operatorname {Ln} \frac{z+i}{z+1} in the plane with a cut along the line

{z:∣z∣=1,−π2⩽arg⁡z⩽π} \left\{ z:\left|z\right| = 1,-\frac{\pi }{2} \leqslant \arg z \leqslant \pi \right\}

such that h(0)=iπ2h(0) = i \frac{\pi }{2}. Expand h(z)h(z) in a Laurent series in powers of zz in a neighborhood of the point z=∞z = \infty.

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Let h(z)h(z) be a regular branch of the multivalued function

Ln⁡z−iz+i−1z−2 \operatorname {Ln} \frac{z-i}{z+i}-\frac{1}{z-2}

in the plane with the point z=2z = 2 removed and a cut along the semicircle {z:∣z∣=1,Re(z)⩾0}\left\{ z:\left|z\right| = 1, \mathrm{Re}\left(z\right) \geqslant 0\right\}, such that h(0)=−iπ+12h(0) = -i \pi +\frac{1}{2}. Expand h(z)h(z) in a Laurent series in powers of zz in the annulus 1<∣z∣<21 < \left|z\right| < 2.

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Let g(z)g(z) be a regular branch of the multivalued function

{z2(1−z)3}+1z−3i \left\{ \sqrt[3]{z^{2}(1-z)}\right\} +\frac{1}{z-3 i}

in the plane with the point z=3iz = 3 i removed and a cut along the segment [0,1][0,1], such that g(−1)=23−11+3ig(-1) = \sqrt[3]{2}-\frac{1}{1+3 i}. Expand g(z)g(z) in a Laurent series in powers of zz in the annulus 1<∣z∣<31 < \left|z\right| < 3.

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Let h(z)h(z) be a regular branch of the multivalued function Ln⁡(z2−z−6)\operatorname {Ln}\left(z^{2}-z-6\right) in the plane with cuts along the lines {Re(z)=−2,Im(z)⩽0}\left\{ \mathrm{Re}\left(z\right) = -2, \mathrm{Im}\left(z\right) \leqslant 0\right\} and {Re(z)=3,Im(z)⩽0}\left\{ \mathrm{Re}\left(z\right) = 3, \mathrm{Im}\left(z\right) \leqslant 0\right\}, such that h(4)=ln⁡6h(4) = \ln 6. Expand h(z)h(z) in a Taylor series centered at the point z=1z = 1 and find the radius of convergence of this series.

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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(z2−2z)\operatorname {Ln}\left(z^{2}-2 z\right) in the plane with cuts along the positive part of the imaginary axis and along the segment [0,2][0,2] of the real axis, singled out by the condition

Im(f(−1+3i))=−π2 \mathrm{Im}\left(f(-1+\sqrt{3} i)\right) = -\frac{\pi }{2}

Compute f(1+2i)f(1+2 i). Expand f(z)f(z) in a Taylor series centered at the point z=1+2iz = 1+2 i. Find the radius of convergence. Compute the sum of the series at z=−1+3iz = -1+\sqrt{3} i.

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Let f(z)f(z) be a regular branch of the multivalued function {z4+643}\left\{ \sqrt[3]{z^{4}+64}\right\} in the plane with cuts along the rays

{z:z=2−2it;t⩾−1} and {z:z=−2−2it;t⩾−1}, \begin{aligned} & \left\{ z: z = 2-2 i t ; t \geqslant -1\right\} \\ \text{ and } \quad & \left\{ z: z = -2-2 i t ; t \geqslant -1\right\} , \end{aligned}

singled out by the condition f(22)=423f(2 \sqrt{2}) = 4 \sqrt[3]{2}. Compute f(0)f(0). Expand f(z)f(z) in a Taylor series centered at the point z=0z = 0. Find the radius of convergence. Compute the ratio of the sum of the series to f(z)f(z) at z=−52z = -\frac{5}{2}.

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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(z+iz2)\operatorname {Ln}\left(z+i z^{2}\right) in the plane with cuts along the positive part of the real axis and along the segment [0,i][0, i] of the imaginary axis, singled out by the condition

Im(f)(e−iπ/6)=0 \mathrm{Im}\left(f\right)\left(e^{-i \pi / 6}\right) = 0

Compute f(1+i2)f\left(1+\frac{i}{2}\right). Expand f(z)f(z) in a Taylor series centered at the point z=1+i2z = 1+\frac{i}{2}. Find the radius of convergence. Compute the sum of the series at z=e−iπ/6z = e^{-i \pi / 6}.

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Let f(z)f(z) be a regular branch of the multivalued function {z4+46}\left\{ \sqrt[6]{z^{4}+4}\right\} in the plane with cuts along the rays {z:z=i+t;t⩾−1}\left\{ z: z = i+t ; t \geqslant -1\right\} and {z:z=−i+t;t⩽1}\left\{ z: z = -i+t ; t \leqslant 1\right\}, singled out by the condition f(2i)=−2f(\sqrt{2} i) = -\sqrt{2}. Compute f(0)f(0). Expand f(z)f(z) in a Taylor series centered at the point z=0z = 0. Find the radius of convergence. Compute the ratio of the sum of the series to f(z)f(z) at z=−65iz = -\frac{6}{5} i.

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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(1+z2)\operatorname {Ln}\left(1+z^{2}\right) in the plane with a cut along the ray of the imaginary axis [−i,+i∞)[-i,+i \infty ), with Im(f)(−15)=0\mathrm{Im}\left(f\right)\left(-\frac{1}{5}\right) = 0. Expand f(z)f(z) in a Taylor series centered at the point z=1z = 1 and find the radius of convergence. Compute the sum of the series at z=−15z = -\frac{1}{5}.

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Let f(z)f(z) be a regular branch of the multivalued function {9−z2}\left\{ \sqrt{9-z^{2}}\right\} in the plane with a cut along the circular arc {z:∣z−4i∣=5,Im(z)⩾0}\left\{ z:\left|z-4 i\right| = 5, \mathrm{Im}\left(z\right) \geqslant 0\right\}, with f(4i)=5f(4 i) = 5. Expand f(z)f(z) in a Laurent series in powers of zz in a neighborhood of z=∞z = \infty and find the domain of convergence. Compute the sum of the series at z=4iz = 4 i.

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Let f(z)f(z) be a regular branch of the multivalued function Ln⁡(1−z2)\operatorname {Ln}\left(1-z^{2}\right) in the plane with a cut along the ray of the real axis (−∞,1](-\infty , 1], with Im(f)(i5)=0\mathrm{Im}\left(f\right)\left(\frac{i}{5}\right) = 0.

Expand f(z)f(z) in a Taylor series centered at the point z=−iz = -i. Find the radius of convergence. Compute the sum of the series at z=i5z = \frac{i}{5}.

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Let f(z)f(z) be a regular branch of the multivalued function {z2+16}\left\{ \sqrt{z^{2}+16}\right\} in the plane with a cut along the circular arc {z:∣z+3∣=5,Re(z)⩾0}\left\{ z:\left|z+3\right| = 5, \mathrm{Re}\left(z\right) \geqslant 0\right\}, with the principal part of the Laurent series of the function f(z)f(z) in a neighborhood of z=∞z = \infty equal to zz. Expand f(z)f(z) in a Taylor series centered at the point z=0z = 0, find the radius of convergence. Compute the sum of the series at z=3z = 3.

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Let f(z)f(z) be a regular branch of the multivalued function {z2−14}\left\{ \sqrt[4]{z^{2}-1}\right\} in the plane with a cut along the curve

γ={z:z=eit,0⩽t⩽π;z=t,−∞<t⩽−1} \gamma = \left\{ z: z = e^{i t}, 0 \leqslant t \leqslant \pi ; z = t,-\infty < t \leqslant -1\right\}

such that f(0)=eπi/4f(0) = e^{\pi i / 4}. Let S(z)=∑k=0+∞Ak(z−2i)kS(z) = \sum_{k = 0}^{+\infty } A_{k}(z-2 i)^{k} be the function, regular in its disk of convergence, coinciding with f(z)f(z) in a neighborhood of the point z=2iz = 2 i. Find the radius of convergence of the series S(z)S(z) and compute the integral

∮∣z−2i∣=32S(z)(z−34i)2dz \oint _{\left|z-2 i\right| = \frac{3}{2}} \frac{S(z)}{\left(z-\frac{3}{4} i\right)^{2}} d z
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Let f(z)f(z) be a regular branch of the multivalued function {z(z−1)3}\left\{ \sqrt[3]{z(z-1)}\right\} in the plane with cuts along the curves γ1\gamma_{1} and γ2\gamma_{2}, where

γ1={z:z=it,t⩾0}γ2={z:z=1−it,0⩽t⩽2;z=t−2i,−∞<t⩽1} \begin{aligned} & \gamma _{1} = \left\{ z: z = i t, t \geqslant 0\right\} \\ & \gamma _{2} = \left\{ z: z = 1-i t, 0 \leqslant t \leqslant 2 ; z = t-2 i,-\infty < t \leqslant 1\right\} \end{aligned}

such that f(12)=−143f\left(\frac{1}{2}\right) = -\sqrt[3]{\frac{1}{4}}. Let S(z)=∑k=0+∞Ak(z+3i)kS(z) = \sum_{k = 0}^{+\infty } A_{k}(z+3 i)^{k} be the function, regular in its disk of convergence, coinciding with f(z)f(z) in a neighborhood of the point z=−3iz = -3 i. Find the radius of convergence of the series S(z)S(z) and compute the integral

∮∣z+3i∣=52S(z)(z+i)2dz \oint _{\left|z+3 i\right| = \frac{5}{2}} \frac{S(z)}{(z+i)^{2}} d z
?

Let f(z)f(z) be a regular branch of the multivalued function {1−z24}\left\{ \sqrt[4]{1-z^{2}}\right\} in the plane with a cut along the curve

γ={z:z=eit,0⩽t⩽π;z=1+it,0⩽t<∞} \gamma = \left\{ z: z = e^{i t}, 0 \leqslant t \leqslant \pi ; z = 1+i t, 0 \leqslant t < \infty \right\}

such that f(0)=1f(0) = 1. Let S(z)=∑k=0+∞Ak(z−3i)kS(z) = \sum_{k = 0}^{+\infty } A_{k}(z-3 i)^{k} be the function, regular in its disk of convergence, coinciding with f(z)f(z) in a neighborhood of the point z=3iz = 3 i. Find the radius of convergence of the series S(z)S(z) and compute the integral

∮∣z−3i∣=52S(z)(z−34i)2dz \oint _{\left|z-3 i\right| = \frac{5}{2}} \frac{S(z)}{\left(z-\frac{3}{4} i\right)^{2}} d z
?

In the complex plane with cuts along the curves γ1\gamma_{1} and γ2\gamma_{2}, where

γ1={z:z=t,t⩽−2}γ2={z:z=t,−1⩽t⩽0;z=it−1,0⩽t<∞} \begin{aligned} & \gamma _{1} = \left\{ z: z = t, t \leqslant -2\right\} \\ & \gamma _{2} = \left\{ z: z = t,-1 \leqslant t \leqslant 0 ; z = i t-1,0 \leqslant t < \infty \right\} \end{aligned}

consider a regular branch f(z)f(z) of the multivalued function {z(z+2)3}\left\{ \sqrt[3]{z(z+2)}\right\}, such that

f(−32)=−343 f\left(-\frac{3}{2}\right) = -\sqrt[3]{\frac{3}{4}}

Let S(z)=∑k=0+∞Ak(z−3i)kS(z) = \sum_{k = 0}^{+\infty } A_{k}(z-3 i)^{k} be the function, regular in its disk of convergence, coinciding with f(z)f(z) in a neighborhood of the point z=3iz = 3 i. Find the radius of convergence of the series S(z)S(z) and compute the integral

∮∣z−3i∣=52S(z)(z+2−2i)2dz \oint _{\left|z-3 i\right| = \frac{5}{2}} \frac{S(z)}{(z+2-2 i)^{2}} d z
?

References

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[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.
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