Residues at Simple Poles and Higher-Order Poles

Source: J. E. Santos, Cambridge Part IB Complex Methods, Example Sheet 2, Question 7.

Show that if $f(z)$ and $g(z)$ are analytic, and $g$ has a simple zero at $z=z_0$, the residue of $f(z)/g(z)$ at $z=z_0$ is $f(z_0)/g'(z_0)$. In particular, show that $f(z)/(z-z_0)$ has residue $f(z_0)$.
Prove the formula for the residue of a function $f(z)$ that has a pole of order $N$ at $z=z_0$: \[\lim_{z\to z_0}\left\{ \frac{1}{(N-1)!}\frac{d^{N-1}}{d z^{N-1}} \left((z-z_0)^N f(z)\right) \right\}.\tag{1.1}\]

The residue is given by \[\operatorname{Res}\left(\frac{f(z)}{g(z)},\, z=z_0\right) = \lim_{z\to z_0}\frac{(z-z_0)f(z)}{g(z)} = \lim_{z\to z_0}\frac{f(z) + (z-z_0)f'(z)}{g'(z)}=\frac{f(z_0)}{g'(z_0)},\tag{1.2}\] where we have used the fact that $g(z_0)\neq0$ at the pole. Choosing $g(z) = z-z_0$, we find \[\operatorname{Res}\left(\frac{f(z)}{z-z_0},\, z=z_0\right) = f(z_0).\tag{1.3}\]

For a function $f(z)$ that has pole of order $N$ at $z=z_0$, it admits a Laurent expansion of the form \[f(z) = \sum_{n=1}^N\frac{b_n}{(z-z_0)^n} + \sum_{n=0}^\infty a_n(z-z_0)^n.\tag{1.4}\] Multiplying $f(z)$ by $(z-z_0)^N$, we find \[(z-z_0)^Nf(z) = \sum_{n=1}^N b_n (z-z_0)^{N-n} + \sum_{n=0}^\infty a_n(z-z_0)^{N+n}.\tag{1.5}\] It is therefore easy to see that \[\operatorname{Rez}\left(f(z), \, z=z_0\right) = \lim_{z\to z_0}\left\{ \frac{1}{(N-1)!}\frac{d^{N-1}}{d z^{N-1}} \left((z-z_0)^N f(z)\right) \right\}.\tag{1.6}\]