I'm trying to calculate the following integral:
$$ \frac{1}{2\pi i}\int_{\gamma} \frac{\cos(z)}{z^7-2z^6}dz $$
Where $\gamma$ is the path that traverses the circle centered at $1/4$ with radius $1/2$ counter-clockwise.
It seems obvious that I should use Cauchy's integral formula here, but I cannot figure out how.
I can rewrite the integral as:
$$ \frac{1}{2\pi i}\int_{\gamma} \frac{z\cos(z)}{(z-2)}\frac{1}{(z-0)^7}dz $$
From this I find that the function $\frac{1}{(z-0)^7}$ has a pole at $z=0$, which is in $\gamma$, but $\frac{z\cos(z)}{(z-2)}$ has a zero at $z=0$. Finally, $ \frac{z\cos(z)}{(z-2)}$ has a pole at 2.
But from here I'm unsure about how I should proceed. It seems so simple, but I can't seem to find a way forward.
Any tips are appreciated.
If $f(z)=\frac{\cos z}{z-2}$, then\begin{align}\frac1{2\pi i}\int_\gamma\frac{\cos z}{z^7-2z^6}\,\mathrm dz&=\frac1{2\pi i}\int_\gamma\frac{f(z)}{z^6}\,\mathrm dz\\&=\frac{f^{(5)}(0)}{5!}\end{align}Or you can apply the residue theorem.