Compute $\oint\frac{e^{2z}+\sin z}{(z^2+1)^3}dz$, over the curve C

Question

Compute $I = \oint\frac{e^{2z}+\sin z}{(z^2+1)^3}dz$, over the curve $C:|z+i|=r, r\neq2$

So what I understood from my classes I have to find what it's poles are and then apply the residual theorem but I have some problems when trying to determine in what interval $r$ is and use that..

So what I've done so far is:

$z^2+1=0\implies z_{1,2}=\pm i$ and then I have to check for what $r$ are these roots in the $IntC$, right? Then apply the theorem that $I=2\pi i \sum_{z_k\space singular \space point}Rez(f(z), z_k)$

Say for $z_1\in IntC\implies |i+i|< r\leftrightarrow 2 < r$, but for $z_2\in IntC\implies |-i+i|< r\leftrightarrow 0 < r$, how do I deal with this?

And we have $\pm i$ poles of order $3$ for that polynomial.. how do I solve it from here? And how do I prove that the integral is $0$ if the singular points are not in the interior of the curve?

Answer 1

There are two possibilities here:

1. $r<2$: In this case, the only residue that have have to deal with is the residue at $-i$. The residue theorem tells us then that the integral is equal to\begin{align}2\pi i\operatorname{res}_{z=-i}\left(\frac{e^{2z}+\sin z}{(z^2+1)^3}\right)&=\frac{2\pi i}{16}e^{-2 i}\left((6+i)-4 e^{2 i}\sinh1+3e^{2i}\cosh1\right)\\&=\frac{\pi i}8e^{-2 i}\left((6+i)-4 e^{2 i}\sinh1+3e^{2i}\cosh1\right).\end{align}
2. $r>2$: In this case, you will have also to deal with the residue at $i$. Since$$\operatorname{res}_{z=i}\left(\frac{e^{2z}+\sin z}{(z^2+1)^3}\right)=\frac1{16}\left((-6+i)e^{2 i}+4\sinh1-3\cosh1\right)$$the integral will be then equal to (after some manipulations):$$\frac\pi8\left(-\frac{7i}e+(1-6i)e^{-2 i}-(1+6i)e^{2i}+ie\right).$$

Note that the singularity $-i$ is always in the region bounded by the curve.

Answer 2

You are on the right track...

If $r < 2$ the only pole in the interior of $C$ is $z=i$ and the integral is given by

$$ \int_C f(z) dz = 2\pi i Res(f,-i) $$ If $r > 2$ there are two poles inside the curve and the integral is given by

$$ \int_C f(z) dz = 2 \pi i (Res(f,-i) + Res (f,i)) $$