- #1
binbagsss
- 1,326
- 12
Homework Statement
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##
Homework 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
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.