# Complex analysis f'/f , f meromorphic, Laurent series

1. Nov 20, 2016

### binbagsss

1. The problem statement, all variables and given/known data

consider $f$ a meromorphic function with a finite pole at $z=a$ of order $m$.
Thus $f(z)$ has a laurent expansion: $f(z)=\sum\limits_{n=-m}^{\infty} a_{n} (z-a)^{n}$

I want to show that $f'(z)'/f(z)= \frac{m}{z-a} + holomorphic function$

And so where a holomporphic function is one where the laurent expansion above will instead start at $n=0$

2. Relevant equations

So I've tried writing out a few terms in the both $f'(z)$ and $f(z)$ to try and see this, but I don't seem to be going anywhere...

$f(z)=\sum\limits_{n=-m}^{\infty} a_{n} (z-a)^{n}$, for a meromphic function with pole at a of order m
$f(z)=\sum\limits_{n=0}^{\infty} a_{n} (z-a)^{n}$, for a holomorphic function

3. The attempt at a solution

$f(z) = a_{-m} / (t-a)^{m} + a_{-m+1}/(t-a)^{m-1} + a_{-m+2}/(t-a)^{m-2} +...+ a_{-1}/(t-a)+a_{0} + a_{1}(t-a) +...$

$f'(z) = -m a_{-m} (t-a)^{-m-1} +(-m+1)a_{m+1}(t-a)^{-m} + (-m+2) a_{-m+2} (t-a)^{-m+1} + ... + (-a_{-1} (t-a)^{-2} + 0 + a_{1} + 2a_{2}(t-a)+...$

I have no idea how to divide properly.
So I look at the terms with the $a_{-m}$ coefficient and see that $-m a_{-m} (t-a)^{-m-1} / ( a_{-m} / (t-a)^{m} ) = -m / (t-a)$ which looks sort of on track,

however....i get the same when i do this with each of the next coefficients of a , i.e. the $(t-a)^{-1}$

help greatly appreciated, many thanks.

2. Nov 20, 2016

### lurflurf

hint
$$(z-a)^mf(z)=holomorphic function \\ (z-a)^{m+1}f^\prime(z)=holomorphic function \\ (z-a)\frac{f^\prime(z)}{f(z)}=holomorphic function$$