Complex analysis applications-II

A possible solution could be:

• Consider the following rational complex function: $f:\mathbb{C} \rightarrow \mathbb{C}, \quad f(z)=\frac { 1 }{ { (z+1+i) }^{ 2 } }$ we will include the singularities,
• For this function suppose that the function has the following propriety: $\exists M> 0$ , such that for a $|z|$ sufficiently large $\implies \; { |z }^{ 2 }f(z)|\le M$ ,
• We will apply the Residue Theorem on the circles given by the equation: $\Gamma :\;|z|=\sigma +\frac { 1 }{ 2 },\; \sigma \in \mathbb{N}$ , and we get:

$I=\oint _{ \Gamma }{ f(z)\cot { (\pi z) } dz } = \oint _{ \Gamma }^{ }{ \frac { 1 }{ { (z+1+i) }^{ 2 } } \cot { (\pi z) } dz } = \oint _{ \Gamma }^{ }{ \frac { 1 }{ { (z+1+i) }^{ 2 }\tan { (\pi z) } } dz }$

• There the following singularities: $z_{0} = -1-i$ and $z_{k} = k, \; k\in \mathbb{Z}$ .
• Applying Residue Theorem for the integral from above :

$I = \oint _{ \Gamma }^{ }{ \frac { 1 }{ { (z+1+i) }^{ 2 }\tan { (\pi z) } } dz } = 2\pi i Res{(f;z_0)}+2\pi i \sum _{ k=-s }^{ s }{ Res(f;z_{ k }) }$ .

• Now we will compute the residues from above:

$R_1 = Res(f,z_0)=\lim _{ z\rightarrow -1-i }{ \Bigg( {(z+1+i) }^{ 2 }\frac { 1 }{ { (z+1+i) }^{ 2 }\tan { \pi z } } \Bigg)\frac { d }{ dz } }=\lim _{ z\rightarrow -1-i }{ (\cot { (\pi z } ) } \frac { d }{ dz } =\lim _{ z\rightarrow -1-i } \frac { -\pi }{ \sin ^{ 2 }{ (\pi z) } } = \frac { -\pi }{ \sin ^{ 2 }{ (\pi +\pi i) } } = \boxed { \frac { \pi }{ \sinh ^{ 2 }{ \pi } } }.$

$R_2 = Res(f;z_k)= \frac { 1 }{ (z+1+i)^{ 2 }(\tan (\pi z))\frac { d }{ dz } } \Bigg|_{z=k} = \frac { \cos ^{ 2 }{ (\pi z) } }{ \pi (z+1+i)^{ 2 } } \Bigg|_{z=k} = \boxed{ \frac { 1 }{ \pi (k+1+i)^2 } }.$

• So we get that:

$I=2 \pi i \frac { \pi }{ \sinh ^{ 2 }{ \pi } } + 2 \pi i \sum _{ k=-s }^{ s }{ \frac { 1 }{ \pi (k+1+i)^2 } }$

• Now we will make $|z| \rightarrow \infty$ and one can prove that:

$\lim_{|z| \rightarrow \infty} {\oint _{ \Gamma }{ f(z)\cot { (\pi z) } dz }} = 0$

using the function's propriety stated in the beginning.

• In the end, for $\sigma \rightarrow \infty$ we obtain:

$\frac { 2\pi^2 i }{ \sinh^2{(\pi)} } + 2i \sum _{ k= -\infty }^{ \infty }{ \frac { 1 }{ (k+1+i)^2 } } = 0 \implies \sum _{ k= -\infty }^{ \infty }{ \frac { 1 }{ (k+1+i)^2 } } = \frac { -\pi^2 }{ \sinh^2{\pi} } = \boxed{-0.073999}.$