Complex Analysis: Problems with Solutions

Solution:
The function [tex]v[/tex] is defined for [tex]z\ne 0[/tex]. We find [tex]u[/tex] from the Cauchy-Riemann equations:
[tex]u_x=v_y,\qquad u_y=-v_x[/tex]
The partial derivatives of [tex]v[/tex] are
[tex]v_x=-\frac{2xy}{(x^{2}+y^{2})^{2}},\qquad v_y=\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}[/tex]
Integrating [tex]u_x=v_y[/tex] with respect to [tex]x[/tex]:
[tex]u=\int\frac{x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}\,dx=-\frac{x}{x^{2}+y^{2}}+\varphi(y)[/tex]
Then [tex]u_y=\frac{2xy}{(x^{2}+y^{2})^{2}}+\varphi'(y)[/tex], and the second equation [tex]u_y=-v_x=\frac{2xy}{(x^{2}+y^{2})^{2}}[/tex] gives [tex]\varphi'(y)=0[/tex], i.e. [tex]\varphi(y)=C[/tex] is a real constant. Hence
[tex]f=u+iv=-\frac{x-iy}{x^{2}+y^{2}}+C=-\frac{\bar z}{|z|^{2}}+C=-\frac{1}{z}+C[/tex]
From the condition [tex]f(2)=0[/tex]:
[tex]f(2)=-\frac{1}{2}+C=0\Rightarrow C=\frac{1}{2}\Rightarrow f(z)=\frac{1}{2}-\frac{1}{z}[/tex]
Check: [tex]-\frac{1}{z}=\frac{-x+iy}{x^{2}+y^{2}}[/tex] has imaginary part [tex]\frac{y}{x^{2}+y^{2}}[/tex].

Solution:
The integrand [tex]x=\operatorname{Re}z[/tex] is not an analytic function of [tex]z[/tex], so the integral depends on the path and we compute it by a parametrization of [tex]L[/tex]. Put [tex]z=e^{it}[/tex], where [tex]t[/tex] goes from [tex]0[/tex] to [tex]\pi[/tex] (the arc starts at [tex]z=1[/tex]). Then
[tex]x=\cos t,\qquad dz=ie^{it}\,dt[/tex]
[tex]\int_L x\,dz=i\int_0^{\pi}\cos t\,(\cos t+i\sin t)\,dt=i\int_0^{\pi}\cos^{2}t\,dt-\int_0^{\pi}\sin t\cos t\,dt[/tex]
We have
[tex]\int_0^{\pi}\cos^{2}t\,dt=\int_0^{\pi}\frac{1+\cos 2t}{2}\,dt=\frac{\pi}{2}[/tex]
[tex]\int_0^{\pi}\sin t\cos t\,dt=\left[\frac{\sin ^{2}t}{2}\right]_0^{\pi}=0[/tex]
Therefore
[tex]\int_L x\,dz=\frac{\pi i}{2}[/tex]

Solution:
The integrand [tex]\frac{1}{(z-1)^{2}(z+1)^{2}}[/tex] is analytic except at [tex]z=\pm 1[/tex]. The circle [tex]L[/tex] has center [tex]-1-i[/tex] and radius [tex]\sqrt{2}[/tex], and
[tex]|-1-(-1-i)|=1\lt\sqrt{2},\qquad |1-(-1-i)|=\sqrt{5}\gt\sqrt{2}[/tex]
so [tex]z=-1[/tex] lies inside [tex]L[/tex] and [tex]z=1[/tex] lies outside. Write the integrand as [tex]\frac{g(z)}{(z+1)^{2}}[/tex] with [tex]g(z)=\frac{1}{(z-1)^{2}}[/tex], which is analytic inside and on [tex]L[/tex]. By Cauchy's integral formula for the first derivative, with [tex]a=-1[/tex],
[tex]\oint_L\frac{g(z)}{(z-a)^{2}}\,dz=2\pi i\,g'(a)[/tex]
Since [tex]g'(z)=-\frac{2}{(z-1)^{3}}[/tex], we get
[tex]g'(-1)=-\frac{2}{(-2)^{3}}=\frac{1}{4}\Rightarrow\oint_L\frac{dz}{(z^{2}-1)^{2}}=2\pi i\times \frac{1}{4}=\frac{\pi i}{2}[/tex]

Solution:
The hyperbola does not pass through [tex]z=0[/tex], so [tex]w=\frac{1}{z}[/tex] is defined at all its points and [tex]w\ne 0[/tex]. The map [tex]z\mapsto\frac{1}{z}[/tex] is a bijection of [tex]\mathbb{C}\setminus\{0\}[/tex] onto itself, with inverse
[tex]z=\frac{1}{w}=\frac{\bar w}{|w|^{2}}=\frac{u-iv}{u^{2}+v^{2}}\Rightarrow x=\frac{u}{u^{2}+v^{2}},\quad y=-\frac{v}{u^{2}+v^{2}}[/tex]
Therefore
[tex]x^{2}-y^{2}=\frac{u^{2}-v^{2}}{\left(u^{2}+v^{2}\right)^{2}}[/tex]
Hence, for [tex]w\ne 0[/tex], the point [tex]z=\frac{1}{w}[/tex] lies on the hyperbola [tex]x^{2}-y^{2}=1[/tex] if and only if
[tex]\left(u^{2}+v^{2}\right)^{2}=u^{2}-v^{2}[/tex]
So the image of the hyperbola is exactly the set of points [tex]w\ne 0[/tex] of the lemniscate of Bernoulli. The point [tex]w=0[/tex], which lies on the lemniscate, is not attained: it corresponds to [tex]z=\infty[/tex] (when [tex]z\to\infty[/tex] along a branch of the hyperbola, [tex]w\to 0[/tex]).