Finding the order of a pole of a complex function

Assume that $lim_{z rightarrow z_0} (z-z_0)^m f(z) < infty$. This means that $(z-z_0)^m f(z)$ has at most (or worst if you will) a removable singularity at $z_0$. Make sure you understand why this is or prove it!

Consider tha Laurent series of $f$ around $z_0$:

$$f(z) = sum_{n = -infty}^{infty} a_n (z-z_0)^n$$

$$(z-z_0)^mf(z) = (z-z_0)^m sum_{n = -infty}^{infty} a_n (z-z_0)^n = sum_{n = -infty}^{infty} a_n (z-z_0)^{n+m}$$

Since, as we commented earlier, this function has at most a removable singularity at $z_0$, $a_n = 0$ for all $n+m < 0 implies n < -m$, i.e.

$$(z-z_0)^m f(z) = sum_{n=-m}^{infty} a_n(z-z_0)^{n+m}.$$

We divide by $(z-z_0)^m$ and get

$$f(z) = sum_{n=-m}^{infty} a_n (z-z_0)^n,$$

which shows that $z_0$ is a pole of order $m$.