Evaluate $\int_{\gamma}\frac{z^5}{z^6-1}$ 
Evaluate $$\int_{\gamma}\frac{z^5}{z^6-1}$$ where $\gamma=\{(x,y)\in\mathbb{R^2}\,|\,x^2+4y^2=16\}$.

How should I approach this? first to find the singularities by $z^6-1=0\iff z^6=1$ 
And then to plug each singularity in  $\{(x,y)\in\mathbb{R^2}x^2+4y^2=16\}$?

Since the curve $\gamma$ (ellipse centered at the origin with semiaxis $2$ and $4$) contains all the finite poles of the rational function $\frac{z^5}{z^6-1}$ ($6$ poles on the unit circle centered at the origin), it follows that
\begin{align*}\int_{\gamma}\frac{z^5}{z^6-1}dz&=-2\pi i\mbox{Res}\left(\frac{z^5}{z^6-1},\infty\right)=2\pi i\mbox{Res}\left(\frac{1/z^5}{z^2(1/z^6-1)},0\right)\\
&=2\pi i\mbox{Res}\left(\frac{1}{z(1-z^6)},0\right)=
2\pi i\mbox{Res}\left(\frac{1+z^6+o(z^6)}{z},0\right)=2\pi i
\end{align*}
where we used the notion of residue at infinity.
