TITLE: Residue of $f'/f$ is equal to $m$.
QUESTION [6 upvotes]: So I know that if $f$ is analytic and has a zero of order $m$ at its center, then $f'$ has a zero of order of $m-1$, which can be easily proved.  
However, I'm not sure, which is stated in the question, how this is related to residues, and how to prove that the residue of $f'/f$ is equal to $m$ if $f$ is analytic and has a zero of order $m$.  Any help would be much appreciated, thank you. 

If $f$ is analytic in $|z - z_0| < R$ and has a zero of order $m$ at $z_0$, show that $$Res\left( \frac{f'}{f} ; z_0 \right) = m.$$

REPLY [5 votes]: Since $f$ is analytic with a zero of order $m$ at $z_0$, we can write $f(z)=(z-z_0)^mg(z)$, where $g$ is analytic and $g(z_0)\ne0$.
Then, we have
$$\begin{align}
\frac{f'(z)}{f(z)}&=\frac{m(z-z_0)^{m-1}g(z)+(z-z_0)^mg'(z)}{(z-z_0)^mg(z)}\\\\
&=\frac{m}{z-z_0}+\frac{g'(z)}{g(z)}
\end{align}$$
Can you finish now?