Recovering an analytic function from its real or imaginary part

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Let u:R2→Ru : \mathbb {R}^{2} \to \mathbb {R} be the function defined by u(x,y)=x3−3xy2u(x,y) = x^{3} - 3xy^{2}. Show that uu is harmonic and find v:R2→Rv : \mathbb {R}^{2} \to \mathbb {R} such that the function f:C→Cf : \mathbb {C} \to \mathbb {C} defined by

f(x+iy)=u(x,y)+iv(x,y) f(x+iy) = u(x,y) + iv(x,y)

is analytic.

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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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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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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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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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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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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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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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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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Find the analytic function of zz that vanishes for z=0z = 0 and has the real part

u=x(1+x2+y2)1+2x2−2y2+(x2+y2)2. u = \frac{x(1+x^{2}+y^{2})}{1+2x^{2}-2y^{2}+(x^{2}+y^{2})^{2}}.
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We denote by aa and bb, a<ba < b, two fixed real numbers, by zz a variable point in the half-plane Im(z)>0\mathrm{Im}\left(z\right) > 0 and by ω\omega the variable angle under which the interval [a,b][a,b] is seen from the point zz. If possible find an analytic function the real part of which is ω\omega.

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We assume that the function

f(z)=a0+a1z+a2z2+⋯+anzn+⋯is regular in the disk∣z∣<R. f(z) = a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n}+\cdots \quad \text{is regular in the disk} \quad \left|z\right|<R.

Express the coefficients a1,a2,…,an,…a_{1},a_{2},\ldots ,a_{n},\ldots in terms of the real and the imaginary part resp.of f(z)f(z) on the circle ∣z∣=r\left|z\right|=r, 0<r<R0<r<R.

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In the following problems, one of the pair of conjugate harmonic functions uu or vv is given. Find the second function of the pair.

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(1)

u=xyu = x y;

(2)

u=x2−y2+2xyu = x^{2}-y^{2}+2 x y;

(3)

v=ysin⁡xch⁡y+xsh⁡ycos⁡xv = y \sin x \operatorname {ch} y+x \operatorname {sh} y \cos x;

(4)

u=rφcos⁡φ+rln⁡rsin⁡φ(z=x+iy=reiφ)u = r \varphi \cos \varphi +r \ln r \sin \varphi \quad \left(z = x+i y = r e^{i \varphi }\right).

Let P(z)P(z) be a polynomial. Set

u(x,y)=Re(P(x+iy));v(x,y)=Im(P(x+iy)). \begin{aligned} & u(x, y) = \mathrm{Re}\left(P(x+i y)\right) ; \\ & v(x, y) = \mathrm{Im}\left(P(x+i y)\right). \end{aligned}

Prove that the formulas hold

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(1)

P(z)=2u(z2,z2i)−P(0)‾P(z) = 2 u\left(\frac{z}{2}, \frac{z}{2 i}\right)-\overline{P(0)};

(2)

P(z)=2iv(z2,z2i)+P(0)‾P(z) = 2 i v\left(\frac{z}{2}, \frac{z}{2 i}\right)+\overline{P(0)}.

Recover the regular function f(z)f(z) from the condition

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(1)

Re(f(z))=x3−6x2y−3xy2+2y3,f(0)=i\mathrm{Re}\left(f(z)\right) = x^{3}-6 x^{2} y-3 x y^{2}+2 y^{3}, f(0) = i;

(2)

Re(f(z))=xsin⁡xch⁡y−ysh⁡ycos⁡x,f(0)=0\mathrm{Re}\left(f(z)\right) = x \sin x \operatorname {ch} y-y \operatorname {sh} y \cos x, \quad f(0) = 0;

(3)

Im(f(z))=ych⁡xcos⁡y+xsin⁡ysh⁡x,f(0)=1\mathrm{Im}\left(f(z)\right) = y \operatorname {ch} x \cos y+x \sin y \operatorname {sh} x, f(0) = 1;

(4)

Im(f(z))=ycos⁡xch⁡y−xsin⁡xsh⁡y,f(0)=2\mathrm{Im}\left(f(z)\right) = y \cos x \operatorname {ch} y-x \sin x \operatorname {sh} y, f(0) = 2;

(5)

Im(f(z))=xsin⁡ych⁡x+ysh⁡xcos⁡y,f(0)=0\mathrm{Im}\left(f(z)\right) = x \sin y \operatorname {ch} x+y \operatorname {sh} x \cos y, f(0) = 0;

(6)

Re(f(z))=xexcos⁡y−(y+1)exsin⁡y,f(0)=i\mathrm{Re}\left(f(z)\right) = x e^{x} \cos y-(y+1) e^{x} \sin y, \quad f(0) = i;

(7)

∣f(z)∣=(x2+y2)ex\left|f(z)\right| = \left(x^{2}+y^{2}\right) e^{x};

(8)

arg⁡f(z)=2xy\arg f(z) = 2 x y;

(9)

∣f(z)∣=rer2cos⁡2φ(z=reiφ)\left|f(z)\right| = r e^{r^{2} \cos 2 \varphi } \quad \left(z = r e^{i \varphi }\right);

(10)

arg⁡f(z)=φ+rsin⁡φ(z=reiφ)\arg f(z) = \varphi +r \sin \varphi \left(z = r e^{i \varphi }\right).

References

[deSi04] de Souza P.N., Silva J. Berkeley Problems in Mathematics. 3rd ed., Springer, 2004.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.