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

  1. Nov 20, 2016 #1
    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. jcsd
  3. Nov 20, 2016 #2

    lurflurf

    User Avatar
    Homework Helper

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