Value of $I$ where $I=\frac{1}{2\pi i} \oint_{\gamma}z^7\cos(\frac{1}{z^2})dz$

Question

Given a counter $\gamma=\{z\in\mathbb{C}:|z|=2\} $ Find the Value of $I$ where

$$I=\frac{1}{2\pi i} \oint_{\gamma}z^7\cos(\frac{1}{z^2})dz$$

Solution I tried-I try to apply the rule that sum of all the residues=negative times the residue at $\infty$ $$Res(f,\infty)+Res(f,0)=0$$ $$Res(f,0)=-Res(f,\infty)$$ but we know that $$Res(f,\infty)=-Res(\frac{1}{z^2}f(\frac{1}{z}),0)$$ after aplying this to my queston i get $$Res(f,\infty)=-Res(\frac{\cos(z^2)}{z^9})$$

now this is even more complicated

Please help

Answer 1

Since$$\cos\left(\frac1{z^2}\right)=1-\frac1{2z^4}+\frac1{4!z^8}-\cdots,$$you have$$z^7\cos\left(\frac1{z^2}\right)=z^7-\frac12z^3+\frac1{4!z}-\cdots$$and therefore$$\operatorname{res}_{z=0}\left(z^7\cos\left(\frac1{z^2}\right)\right)=\frac1{4!}=\frac1{24}.$$So, $I=\dfrac1{24}$.

Answer 2

It's not more complicated. If you expand the residuand at $z=0$, you get $$\frac{\cos(z^2)}{z^9} = \frac{1}{z^9} \left(1-\frac 1 2z^4+ \frac 1 {4!}z^8-\frac 1 {6!} z^{12} + \cdots \right) = \frac 1 {z^9} - \frac 1 {2z^5} + \frac 1{4! z} -\frac{z^3}{6!} + \cdots$$ What's the residue at $z=0$?