I have the following integral and I know that is $0$ but why? $$\int _T \frac{1}{(z^2+4)(z^2+16)^n)}dz$$ where $n$ is natural and $T$ is the triangle in $(-i,1,-1)$ and $z$ is complex number
I know that is $0$ because the function inside integral it doesn't cancel in any point inside the triangle
Hint: By Cauchy's theorem, if $f(z)$ is analytic on the region bounded by the closed curve $\gamma$, then $\oint_{\gamma}f(z)\operatorname dz=0$.