# Why are the poles of $\frac{f'}{f}$ corresponding to the zeros and poles of $f$?

Why are the poles of $\frac{f'}{f}$ corresponding to the zeros and poles of $f$?

Suppose that $a$ is a zero of $f$ of order $m$, then $\frac{f'}{f}$ has a pole at $a$ of residue $m$, suppose that $b$ is a pole of $f$ of order $n$, then $\frac{f'}{f}$ has a zero at $b$ of residue $-n$. Why are these true?

• Write $f(z) = (z-a)^k\cdot g(z)$ in a neighbourhood of $a$ where $g$ is holomorphic, with $g(a) \neq 0$. – Daniel Fischer Dec 10 '16 at 18:44
• How to get the residue of it? – z.z Dec 10 '16 at 18:47
• If you use that form for $f$, what representation do you get for $\frac{f'(z)}{f(z)}$ near $a$? – Daniel Fischer Dec 10 '16 at 18:48
• I see, but how about poles of $f(z)$ also being the poles of $\frac{f'}{f}$ – z.z Dec 11 '16 at 5:03
• If $f$ has a pole at $z_0$, expand $f$ as a Laurent series in a disk centered at $z_0$ with no poles besides $z_0$. – Ethan Alwaise Dec 11 '16 at 6:13