I was doing the following problem:
$f$ is an analytic function defined in a neighborhood of $0$, and $f'(0)\neq 0$. Show that $$\frac{1}{f'(0)}=\frac{1}{2\pi i}\int_{|z|=r}\frac{1}{f(z)-f(0)}dz,$$ provided that $r>0$ is sufficiently small.
I was trying to use the Cauchy integral formula for the inverse of $f$, but it doesn't seem to give the desired integral. Any help would be appreciated.
The function
$$g(z)=\begin{cases} \frac{z}{f(z)-f(0)} &,z\ne0\\\\\frac1{f'(0)}&,z=0 \end{cases} $$
is analytic on $|z|<r$ for sufficiently small values of $r$.
Hence, Cauchy's Integral Formula guarantees that for any such sufficiently small $r$
$$\begin{align} \frac1{2\pi i}\oint_{|z|=r}\frac{1}{f(z)-f(0)}\,dz&=\frac1{2\pi i}\oint_{|z|=r}\frac{g(z)}{z}\,dz\\\\ &=g(0)\\\\ &=\frac1{f'(0)} \end{align}$$