Residue for quotient of functions Let $f, g$ be holomorphic functions on a disk $\mathbb{D}(z_0,r)$ centered at $z_0$ and of radius $r>0$. Suppose $f$ has a simple zero at $z_0$. I want to find an expression for $Res(g/f, z_0)$. But I'm not sure what this expression should look like.
Here's my guess:
Since $f$ has a simple zero at $z_0$, $\exists h(z)$ holomorphic on $\mathbb{D}(z_0,r)$ such that $f(z)=(z-z_0)h(z)$ and $h(z_0)\ne 0$. So that we can represent $g/f = \frac{g(z)}{(z-z_0)h(z)}$, where we observe that $z_0$ is a pole of order 1 of $g/f$. This implies that we can express $$g/f=\frac{a_{-1}}{z-z_0}+\sum\limits_{n=0}^\infty a_n(z-z_0)^n$$
Hence, $$a_{-1}=\frac{g(z)}{f(z)}(z-z_0)-\sum\limits_{n=0}^\infty a_n(z-z_0)^{n+1}$$
Does this look like a correct approach? I think that this expression is too general because of the infinite series on the right-hand side. Is there a clue I'm missing?
Update:
Another approach might be this:
$$g/f = \frac{g(z)}{(z-z_0)h(z)}=\frac{c_0+c_1(z-z_0)+\dots}{d_1(z-z_0)+\dots}=\frac{c_0}{d_1(z-z_0)+\dots}\\
+\frac{c_1(z-z_0)+\dots}{d_1(z-z_0)+\dots}=\frac{a_{-1}}{z-z_0}+\sum a_n(z-z_0)^n$$
But what next?

Here's what I've finally come up with, with credit to the answer by Zaid Alyafeai.
Since $f$ has a simple zero at $z_0$, we can express it as $f(z)=(z-z_0)h(z)$, where $h(z)$ is holomorphic on $\mathbb{D}(z_0, r)$ and $h(z)\ne 0$ on this set, so that $1/h(z)$ is holomorphic on $\mathbb{D}$ and has Taylor series $1/h(z)=\sum\limits_{k=0}^\infty a_k(z-z_0)^k$. Thus $\frac{1}{f}= \frac{1/h(z)}{z-z_0}=\sum\limits_{k=0}^\infty a_k(z-z_0)^{k-1}=\frac{a_0}{z-z_0}+O(1)$. $g(z)$ is holomorphic on $\mathbb{D}$, so it has Taylor series $g(z)=\sum\limits_{k=0}^\infty b_k (z-z_0)^k=b_0+O[(z-z_0)^2]$. So that $$\frac{g}{f}=\frac{b_0a_0}{z-z_0}+O(1)$$
where $b_0=g(z_0)$. Now, $f'(z)=h'(z)(z-z_0)+h(z)$, and $1/f'(z_0)=1/h(z_0)=a_0$. Hence,
$$Res(\frac{g}{f}, z_0)=\frac{g(z_0)}{f'(z_0)}$$
