Residue Theorem for function quotients. let $G$ be an open disc centered around $z_0$ of radius $r$.
Let $f(z),g(z)$ be holomorphic functions on $G$. such that $f(z)$ has a simple zero at $z_0$.
Find an expression for the residue of $\frac{g(z)}{f(z)}$ at $z_0$.
When I looked this up, I found that the answer is actually $\frac{g(z_0)}{f'(z_0)}$.
I'm currently stuck at this step:
$$\int_{\gamma} \frac{g(z)}{f(z)}\, dz = 2{\pi}i \text{Res}\left(\frac{g(z)}{f(z)}, z_0\right)$$
where ${\gamma}$ is a closed curve in $G$ that wraps around $z_0$
I'm not too sure how to continue. I tried to pull out a $g(z_0)$ by using the cauchy integral formula but I don't think it will work.
 A: Here's a (in my opinion simpler) way to deduce the formula without power series:
$$
\operatorname*{Res}_{z=z_0} g/f = \lim_{z\to z_0} (z-z_0) \cdot \frac{g(z)}{f(z)} = \lim_{z\to z_0} g(z) \frac{z-z_0}{f(z)-f(z_0)} = \frac{g(z_0)}{f'(z_0)}
$$
since $f(z_0) = 0$.
A: Since both $f$ and $g$ are analytic on $G$, and $f$ has a simple zero at $z_0$, we can write 
$$f(z)=\sum_{n=1}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}$$
and 
$$g(z)=\sum_{n=0}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}$$
Then, the ratio $g(z)/f(z)$ has a simple pole at $z_0$ with 
$$\begin{align}
\frac{g(z)}{f(z)}&=\frac{\sum_{n=0}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{\sum_{n=1}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}}\\\\
&=\frac{g(z_0)+\sum_{n=1}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{f'(z_0)(z-z_0)+\sum_{n=2}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^n}{n!}}
\end{align}$$
The residue is simply the limit 
$$\begin{align}
\lim_{z\to z_0}\left(\frac{(z-z_0)g(z)}{f(z)}\right)&=\lim_{z\to z_0}\left(\frac{g(z_0)+\sum_{n=1}^\infty \frac{g^{(n)}(0)\,(z-z_0)^n}{n!}}{f'(z_0)+\sum_{n=2}^\infty \frac{f^{(n)}(z_0)\,(z-z_0)^{n-1}}{n!}}\right)\\\\
&=\frac{g(z_0)}{f'(z_0)}
\end{align}$$
as was to be shown! 
