Assigning a removable singularity a value.

Question

I am having trouble with the following question:

Given a function $h(z)=\frac{f(z)}{g(z)}$, where $f$ is holomorphic and has a zero at $z_0$ of order $n$, while $g$ also is holomorphic with a zero at $z_0$ of order $m$, show that for $n\geq m$, the function $h$ has a removable singularity at $z_0$, and that $$\lim_{z\rightarrow z_0}h(z) = \frac{f^{(m)}(z_0)}{g^{(m)}(z_0)}.$$

Any tips?

Answer 1

Following these steps will help.

1. You can use power series expansion at $z_0$. Why?
2. The coefficients of a term $(z-z_0)^k$ is $a_k=\frac{f^{(k)}(z_0)}{k!}$. How?
3. As a result you will obtain $f(z)=(z-z_0)^m(a_m+O(z-z_0))$. Check this.
4. Conclude.