Evaluate $\int_{\gamma}\frac{z^5}{z^6-1}$

Question

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 $\{(x,y)\in\mathbb{R^2}x^2+4y^2=16\}$?

Answer 1

Apply the argument principle: define $f(z)=z^6-1$. Then your integral is $\frac16\int_\gamma\frac{f'(z)}{f(z)}\,\mathrm dz$. By the argument principle, this is equal to $\frac16\times2\pi i$ times the number of zeros minus the number of poles of $f$ in the region of $\mathbb C$ enclosed by $\gamma$. Since there are six zeros and no poles there, the answer is $2\pi i$.

Answer 2

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.