The problem is:
Let $\gamma$ be the circle of radius $R$ centered at $0$. Let $m$, $n$ be positive integers. Prove that, as $R$ goes to infinity, $\int _\gamma\frac{z^m}{z^n+1}dz=O(R^{m+1-n})$. And deduce that if $n\geq m+2$, then the above integral goes to $0$ as $R$ goes to infinity.
My thoughts: Since $\int _\gamma f(z)dz=\int_{a}^{b}f(\gamma(t))\gamma'(t)dt $ and we have in this problem, $\gamma(t)=Re^{it}$, $a=0$, $b=2\pi$. So
$\mathbf{\int _\gamma\frac{z^m}{z^n+1}dz=\int_{0}^{2\pi}\frac{iR^{m+1}e^{(m+1)it}}{R^ne^{nit}+1} dt}$. But how to evaluate this integral? How to proceed?
Thanks in advance!
Use the fact that,
$$ \vert \int_\gamma f(z) dz \vert \leq \int_\gamma \vert f(z) \vert dz \leq 2\pi R \text{ max}(\vert f(z)\vert)_{z\in \gamma} $$
$$ 2\pi R \vert f(z) \vert =2\pi R \frac{\vert z^m\vert}{\vert 1+z^n \vert} \leq 2\pi R \frac{R^m}{\vert 1-R^n\vert}$$