Calculate \begin{equation*} \int_{\Gamma}\frac{1}{z^4+16}dz, \end{equation*} where $\Gamma :|z-i|=\frac{1}{9}$. I have asked I similar question to this but I still do not understand.... when I find the roots of $z^4$+$16$ none of them are in the circle so is the answer just $0$?
I cannot find any similar questions to this to try to work from so I am very confused, please help me
Thanks
Yes, if none of them are in the circle the answer is just $0$.
Cauchy's theorem says that if the function is holomorphic in the inside of $\Gamma$, and the inside is simply connected, then the integral is $0$.
THis is precisely your case.
The problem with curves that go around $0$ with functions like $\frac 1z$ is that the function is holomorphic in the circle except $0$ (a singularity!) which makes the region in which $\frac 1z$ is holomorphic not simply connected.
If you do not have singularities, though, the answer will always be $0$