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
Best Answer
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, $I=\dfrac1{24}$.