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?

Suppose that $z_0$ is a simple zero of the fraction 
$$\mathrm {Res}(g/f,z_0) = \lim_{z\to z_0}(z-z_0) \frac {g(z)}{f(z)}=\frac {g(z_0)}{f'(z_0)} \tag{1}$$

Alternatively 
Since $f$ has a simple zero at $z_0$
$$\frac{1}{f} = \frac{a_{-1}}{(z-z_0)}+\sum_{k=0}^\infty a_k (z-z_0)^k$$
Suppose that $g$ is analytic in a nbhd of $z_0$
$$g(z) = \sum_{k=0}^\infty b_k(z-z_0)^k$$
Then 
$$G(z)=\frac{g(z)}{f(z)} = \left( \sum_{k=0}^\infty b_k(z-z_0)^k \right)\left( \frac{a_{-1}}{(z-z_0)}+\sum_{k=0}^\infty a_k (z-z_0)^k\right)$$
Finally we have the residue at $z_0$
$$\mathrm {Res}(G(z),z_0) = a_{-1}b_0 \tag{2}$$
