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

In summary, the problem is asking to show that for a meromorphic function with a finite pole at z=a of order m, the derivative of f(z) divided by f(z) is equal to m/(z-a) plus a holomorphic function. The initial attempt at a solution involved expanding the Laurent series for f(z) and f'(z), but the key insight is to use the equations given in the Homework Equations section to simplify the problem. By multiplying both sides of the equation by (z-a)^m, it can be shown that f(z) and f'(z) are both holomorphic functions. Then, using the equation for the quotient of two holomorphic functions, it can be shown that the derivative of f(z
  • #1
binbagsss
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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.
 
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  • #2
hint
$$(z-a)^mf(z)=holomorphic function \\
(z-a)^{m+1}f^\prime(z)=holomorphic function \\
(z-a)\frac{f^\prime(z)}{f(z)}=holomorphic function$$
 

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

What is complex analysis?

Complex analysis is a branch of mathematics that deals with the study of complex numbers and functions. It involves the analysis of functions that are defined on complex numbers, as well as their derivatives and integrals.

What is f'/f in complex analysis?

f'/f, also known as the logarithmic derivative, is a mathematical expression used in complex analysis to describe the behavior of a function f(z) in terms of its derivative f'(z). It is defined as f'(z)/f(z) and is often used to study the growth and decay of complex functions.

What does it mean for a function to be meromorphic?

A function f(z) is said to be meromorphic if it is analytic everywhere except for a finite number of isolated singularities. In other words, it is a function that is both analytic and has poles in the complex plane.

What is a Laurent series in complex analysis?

A Laurent series is a representation of a complex function as an infinite sum of terms involving powers of (z-z0), where z0 is a point in the complex plane. It is a generalization of a Taylor series, which only includes non-negative powers of (z-z0).

How is complex analysis used in real-world applications?

Complex analysis has various applications in physics, engineering, and other areas of science. It is used to solve problems involving electric and magnetic fields, fluid flow, and heat transfer. It is also used in signal processing, image processing, and control theory.

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