Let $t$ be a complex variable and $t$ describes a small circle round the origin (inside the circle $|t|=1$).
Let $u$ be defined by $u=\frac{2t}{1-t^2}.$
The author then said that, $u$ also describes a small circle. I cannot see this. Note that $t^2$ is still complex. (I don't know if he means $|t|^2$ and this is a quite old text.) Any help on this fact will be appreciated.
Next, he wrote that,
$$z^m=\frac{m!}{2\pi i}\int_{\Gamma_u} u^{m-1} \exp(z/u)du,$$ where $z$ is a complex variable and $m$ is an integer and $\Gamma_u$ is the small circle which $u$ describes.
I know the Cauchy integral formula (the one for derivatives). But the integrand doesn't contain the factor $\frac{1}{(u-z)^{m+1}}$.
Can anyone help me understand why $\Gamma_u$ is also a small circle and why the equality above holds.
Many thanks.
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,$$\frac1{2\pi i}\int_{\Gamma_u}u^{m-1}\exp\left(\frac zu\right)\,\mathrm du=\frac{z^m}{m!}.$$