Complex Analysis: Difficult Problems with Solutions

Solution:
A linear fractional map sends lines and circles to lines and circles. The line [tex]l[/tex] (that is, [tex]y=-x[/tex]) does not pass through the pole [tex]z=2[/tex], so its image does not contain [tex]w=\infty[/tex] and is a circle [tex]k[/tex].
By the symmetry principle, points symmetric with respect to [tex]l[/tex] go to points symmetric with respect to [tex]k[/tex]. The pole [tex]z=2[/tex] goes to [tex]w=\infty[/tex], and the point symmetric to [tex]\infty[/tex] with respect to a circle is its center. The reflection in the line [tex]y=-x[/tex] maps [tex](x,y)[/tex] to [tex](-y,-x)[/tex], so the point symmetric to [tex]z=2[/tex] is [tex]z=-2i[/tex]. Hence
[tex]w_0=w(-2i)=\frac{2-2i}{-2-2i}=-\frac{1-i}{1+i}=-\frac{(1-i)^{2}}{2}=i[/tex]
The point [tex]z=0\in l[/tex] goes to [tex]w=-1\in k[/tex], so
[tex]R=|-1-w_0|=|-1-i|=\sqrt{2}[/tex]
Check with another point of [tex]l[/tex]: [tex]z=1-i[/tex] gives [tex]w=\frac{3-i}{-1-i}=-1+2i[/tex], and [tex]|-1+2i-i|=|-1+i|=\sqrt{2}[/tex].

Solution:
Put [tex]w=z-i[/tex]. The only singular point [tex]z=0[/tex] corresponds to [tex]w=-i[/tex], [tex]|w|=1[/tex], so the function is analytic in the region [tex]|w|\gt 1[/tex] and has a Laurent expansion there. Write [tex]z=w+i=w\left(1+\frac{i}{w}\right)[/tex]; for [tex]|w|\gt 1[/tex] we have [tex]\left|\frac{i}{w}\right|\lt 1[/tex].
Differentiating the geometric series term by term, for [tex]|q|\lt 1[/tex] we get
[tex]\frac{1}{1+q}=\sum_{n=0}^{\infty}(-q)^{n}\Rightarrow\frac{1}{(1+q)^{2}}=\sum_{n=0}^{\infty}(n+1)(-q)^{n}[/tex]
With [tex]q=\frac{i}{w}[/tex] we obtain, for [tex]|z-i|\gt 1[/tex],
[tex]\frac{1}{z^{2}}=\frac{1}{w^{2}}\times \frac{1}{\left(1+\frac{i}{w}\right)^{2}}=\sum_{n=0}^{\infty}\frac{(n+1)(-i)^{n}}{w^{n+2}}=\sum_{n=0}^{\infty}\frac{(n+1)(-i)^{n}}{(z-i)^{n+2}}[/tex]
The first terms are
[tex]\frac{1}{z^{2}}=\frac{1}{(z-i)^{2}}-\frac{2i}{(z-i)^{3}}-\frac{3}{(z-i)^{4}}+\ldots[/tex]
(In the disc [tex]|z-i|\lt 1[/tex] one gets instead the Taylor series [tex]-\sum_{n=0}^{\infty}(n+1)i^{n}(z-i)^{n}[/tex], which does not converge in the given region.)

Solution:
Factor the denominator:
[tex]z^{3}+2z^{2}+2z=z(z^{2}+2z+2)=z(z+1-i)(z+1+i)[/tex]
The numerator does not vanish at [tex]0[/tex], [tex]-1+i[/tex], [tex]-1-i[/tex], so these are simple poles of the integrand. The circle [tex]C[/tex] has center [tex]-1+i[/tex] and radius [tex]\frac{3}{2}[/tex]; the pole [tex]-1+i[/tex] is its center, and for the other two poles
[tex]|0-(-1+i)|=|1-i|=\sqrt{2}\lt\frac{3}{2},\qquad |-1-i-(-1+i)|=|-2i|=2\gt\frac{3}{2}[/tex]
Hence [tex]0[/tex] and [tex]-1+i[/tex] lie inside [tex]C[/tex], and [tex]-1-i[/tex] lies outside. The residues at the two simple poles inside are (using [tex](-1+i)^{2}=-2i[/tex])
[tex]\operatorname{Res}_{z=0}=\left.\frac{z^{2}+4}{z^{2}+2z+2}\right|_{z=0}=\frac{4}{2}=2[/tex]
[tex]\operatorname{Res}_{z=-1+i}=\left.\frac{z^{2}+4}{z(z+1+i)}\right|_{z=-1+i}=\frac{4-2i}{2i(-1+i)}=\frac{4-2i}{-2-2i}=\frac{-1+3i}{2}[/tex]
By the residue theorem
[tex]\oint_C\frac{(z^{2}+4)\,dz}{z^{3}+2z^{2}+2z}=2\pi i\left(2+\frac{-1+3i}{2}\right)=\pi i(3+3i)=3\pi(i-1)[/tex]

Solution:
The integrand is even, so
[tex]\int_{0}^{\infty}\frac{2x^{2}+1}{x^{4}+4x^{2}+3}\,dx=\frac{1}{2}\int_{-\infty}^{\infty}f(x)\,dx,\qquad f(z)=\frac{2z^{2}+1}{(z^{2}+1)(z^{2}+3)}[/tex]
The rational function [tex]f[/tex] has no real poles, and the degree of its denominator exceeds that of the numerator by [tex]2[/tex]. Hence for large [tex]R[/tex] on the upper semicircle [tex]|z|=R[/tex] we have [tex]|f(z)|\le\frac{c}{R^{2}}[/tex], the integral over it is at most [tex]\pi R\times \frac{c}{R^{2}}=\frac{\pi c}{R}\to 0[/tex] in absolute value, and [tex]\int_{-\infty}^{\infty}f(x)\,dx[/tex] equals [tex]2\pi i[/tex] times the sum of the residues in the upper half-plane. There [tex]f[/tex] has the simple poles [tex]i[/tex] and [tex]i\sqrt{3}[/tex]:
[tex]\operatorname{Res}_{z=i}f=\left.\frac{2z^{2}+1}{(z+i)(z^{2}+3)}\right|_{z=i}=\frac{-1}{2i\,(-1+3)}=\frac{-1}{4i}=\frac{i}{4}[/tex]
[tex]\operatorname{Res}_{z=i\sqrt{3}}f=\left.\frac{2z^{2}+1}{(z^{2}+1)(z+i\sqrt{3})}\right|_{z=i\sqrt{3}}=\frac{-5}{(-2)\,2i\sqrt{3}}=\frac{5}{4\sqrt{3}\,i}=-\frac{5\sqrt{3}}{12}i[/tex]
Therefore
[tex]\int_{-\infty}^{\infty}f(x)\,dx=2\pi i\left(\frac{i}{4}-\frac{5\sqrt{3}}{12}i\right)=-\frac{2\pi\left(3-5\sqrt{3}\right)}{12}=\frac{\pi\left(5\sqrt{3}-3\right)}{6}[/tex]
and the required integral is half of this:
[tex]\int_{0}^{\infty}\frac{2x^{2}+1}{x^{4}+4x^{2}+3}\,dx=\frac{\pi\left(5\sqrt{3}-3\right)}{12}[/tex]

Solution:
Put [tex]g=f-a[/tex]. It is holomorphic in [tex]G[/tex] and [tex]|g(z)|=r[/tex] for all [tex]z\in G[/tex]. If [tex]r=0[/tex], then [tex]g\equiv 0[/tex], i.e. [tex]f\equiv a[/tex]. Let [tex]r\gt 0[/tex] and write [tex]g=U+iV[/tex] with real functions [tex]U,V[/tex]. Then in [tex]G[/tex]
[tex]U^{2}+V^{2}=r^{2}[/tex]
Differentiating with respect to [tex]x[/tex] and [tex]y[/tex] and using the Cauchy-Riemann equations [tex]V_x=-U_y[/tex], [tex]V_y=U_x[/tex], we get
[tex]\begin{cases} UU_x+VV_x=0\\ UU_y+VV_y=0\end{cases}\ \Rightarrow\ \begin{cases} UU_x-VU_y=0\\ VU_x+UU_y=0\end{cases}[/tex]
This is a homogeneous linear system for [tex]U_x,U_y[/tex] with determinant [tex]U^{2}+V^{2}=r^{2}\ne 0[/tex], hence [tex]U_x=U_y=0[/tex] at every point of [tex]G[/tex], and by the Cauchy-Riemann equations also [tex]V_x=V_y=0[/tex]. Thus
[tex]g'(z)=U_x+iV_x=0,\qquad z\in G[/tex]
A domain is connected, and a function with zero derivative on a connected open set is constant. Hence [tex]g[/tex], and therefore [tex]f=g+a[/tex], is constant. (Another argument: by the open mapping theorem a non-constant holomorphic function maps [tex]G[/tex] onto an open set, but a circle contains no non-empty open subset of [tex]\mathbb{C}[/tex].)