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?

  • $\begingroup$ Write $f(z) = (z-a)^k\cdot g(z)$ in a neighbourhood of $a$ where $g$ is holomorphic, with $g(a) \neq 0$. $\endgroup$ – Daniel Fischer Dec 10 '16 at 18:44
  • $\begingroup$ How to get the residue of it? $\endgroup$ – z.z Dec 10 '16 at 18:47
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    $\begingroup$ If you use that form for $f$, what representation do you get for $\frac{f'(z)}{f(z)}$ near $a$? $\endgroup$ – Daniel Fischer Dec 10 '16 at 18:48
  • $\begingroup$ I see, but how about poles of $f(z)$ also being the poles of $\frac{f'}{f}$ $\endgroup$ – z.z Dec 11 '16 at 5:03
  • $\begingroup$ 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 $z_0$. $\endgroup$ – Ethan Alwaise Dec 11 '16 at 6:13

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