An complex integral over circle

Solution 1:

If $t$ belongs to the circle centered at $0$ with radius $r$ (with $r\in(0,1)$), then $t=re^{i\theta}$, and\begin{align}\frac{2t}{1-t^2}&=\frac{2re^{i\theta}}{1-r^2e^{2i\theta}}\\&=\frac{2r}{(1-r^2)\cos(\theta)-(1+r^2)\sin(\theta)i},\end{align}which, for small values of $r$, is close to$$\frac{2r}{\cos(\theta)-\sin(\theta)i}=2re^{i\theta}.$$

On the other hand,\begin{align}u^{m-1}\exp\left(\frac zu\right)&=u^{m-1}\left(1+\frac zu+\frac{z^2}{2u^2}+\cdots+\frac{z^m}{m!u^m}+\cdots\right)\\&=u^{m-1}+zu^{m-2}+\frac12z^2u^{m-3}+\cdots+\frac1{m!}z^mu^{-1}+\cdots,\end{align}and therefore the residue at $0$ of $u^{m-1}\exp\left(\frac zu\right)$ with respect to $u$ is $\frac{z^m}{m!}$. So,$$\frac1{2\pi i}\int_{\Gamma_u}u^{m-1}\exp\left(\frac zu\right)\,\mathrm du=\frac{z^m}{m!}.$$