A few questions on an application of the residue theorem

Question

I have got the following closed curve $\gamma$

and I am supposed to calculate \begin{equation} \int_\gamma\frac{\cos(z)}{z^3(z^2 +1)}dz \end{equation}

with the help of the residue theorem. Now, I've got a few questions

1. $0$, i and -i are the isolated singularities of our given function. As far as I understand it, the winding number of $0$ is $-2$, the winding number of $-$i is $-1$ and the winding number of i is $0$. Is this correct?
2. How can I determine the residue in this case? For a start, I tried finding the Laurent expansion for $z_0=0$, but I have no idea how to rearrange the equation further. I only got as far as \begin{equation} \frac{\cos(z)}{z^3(z^2 +1)}=\frac{1}{z^2+1}\sum\limits_{n=0}^{\infty}(-1)^n\cdot\frac{z^{2n-3}}{(2n)!}. \end{equation} Could someone give me a hint on how to find the Laurent expansion?

Thanks in advance!

Answer 1

1. You are right.
2. Let $f(z)=\frac{\cos(z)}{z^2+1}$. Then\begin{align}\operatorname{res}_{z=0}\left(\frac{\cos(z)}{z^3(z^2+1)}\right)&=\operatorname{res}_{z=0}\left(\frac{f(z)}{z^3}\right)\\&=\frac{f''(0)}{2!}\\&=-\frac32.\end{align}

Answer 2

Your conclusions about the winding numbers and singularities are correct.

At $z=0$, $\cos(z)=1-\frac12z^2++O\!\left(z^4\right)$ and $\frac1{1+z^2}=1-z^2+O\!\left(z^4\right)$. Thus, $$ \frac{\cos(z)}{z^3(1+z^2)}=\frac1{z^3}-\frac32\frac1z+O(z) $$ At $z=\omega$, where $\omega^2+1=0$, the residue is $$ \begin{align} \lim_{z\to\omega}(z-\omega)\frac{\cos(z)}{z^3(z^2+1)} &=\frac{\cos(\omega)}{\omega^3}\lim_{z\to\omega}\frac{z-\omega}{z^2+1}\\ &=\frac{\cos(\omega)}{\omega^3}\frac1{2\omega}\\[3pt] &=\frac{\cosh(1)}2 \end{align} $$