Uniform Limits of Analytic functions, complex analysis question Here's the question I'm trying to answer:
Let $f$ be analytic on an open set $U$, let $z_{0}\in U$ and $f'(z_{0})\neq 0$. Show
\begin{equation*}
    \frac{2\pi i}{f'(z_{0})}=\int _{C}\frac{1}{f(z)-f(z_{0})}dz
\end{equation*}
where C is a small circle centered at $z_{0}$. 
I'm unsure how to start this problem. I've tried some manipulation of the Cauchy Integral formula, but haven't really gotten anywhere. 
Additionally, this problem appears in a section where only one theorem is given, so I'm confident you need to use it but I can't figure out how. Here's the theorem: 
Let $\{f_{n}\}$ be a sequence of analytic functions on an open set $U$, converging uniformly on every compact subset $K\subseteq U$ to a function $f$. Then $f$ is holomorphic. Furthermore, the sequence of derivatives $\{f'_{n}\}$ converges uniformly on every compact subset $K$ to $f'$. 
Any hints would be appreciated. Thanks so much in advance. 
 A: Cauchy integral theorem says that if $f(z)$ is analytic inside a contour then
$\oint_\gamma \frac {f(z)}{z-z_0} \ dz = 2\pi i f(z_0)$
What can we do to get this integral into that form?
$\oint_\gamma \frac {z-z_0}{(z-z_0)(f(z) - f(z_0)} \ dz$
Let $g(z) = \frac {z - z_0}{f(z) - f(z_0)}$
$\oint_\gamma \dfrac {g(z)}{z-z_0} \ dz = 2\pi i g(z_0)$
So what is $g(z_0)$?
$f'(z_0) = \lim_\limits{z\to z_0} \frac {f(z) - f(z_0)}{z-z_0} = \lim_\limits{z\to z_0} \frac 1{g(z)}$
Since $f(z)$ is analyitc and $f'(z_0) \ne 0$ we can say $g(z_0) = \frac {1}{f'(z_0)}$
A: Hint: $1/(f(z)-f(z_0))$ has a simple pole at $z_0$ and nowhere else inside the circle $C$ if $C$ is small enough. Use the residue theorem.
A: Another approach would be to note that $f(z)-f(z_0) = f'(z_0)(z-z_0) + g(z) (z-z_0)^2$, where $g$ is analytic on an open set containing $U$ and for sufficiently small $|z-z_0|$ we have
${1 \over f(z)-f(z_0)} = {1 \over f'(z_0)(z-z_0)} \sum_{k=0}^\infty ( -{ g(z)(z-z_0) \over f'(z_0)} )^k$ and so
$\int_C {dz \over f(z)-f(z_0)} = \int_C {1 \over f(z)'(z-z_0)} dz
= 2 \pi i \operatorname{res}(\eta, z_0)$, where
$\eta(z) = {1 \over f(z)'(z-z_0)}$.
