# How can I find the Laurent series for Cos(1/z) at z=0?

• Crush1986
In summary, the conversation discusses finding the Laurent series of Cos[1/z] at z=0. It is suggested to solve for the series of Cos(u) and then substitute u=z^-1. The Laurent series allows for negative exponents to capture information near the singularity of 1/z.
Crush1986

## Homework Statement

I need to find the Laurent Series of $$Cos[\frac{1}{z}]$$ at z=0

None

## The Attempt at a Solution

I've gone through a lot of these problems and this is one of the last on the problem set. With all the other trig functions it's been just computing their Taylor series, then just simplifying the terms of the problem. Usually, I'd have to do like a geometric series in there as well.

This function though I'm having trouble with because f(0) doesn't exist. I'm not sure what else I can do. Can anyone point me in the right direction?

Thanks.

What if you were asked to solve for the series for ##\cos(u)##? Then let ##u = z^{-1}##?

RUber said:
What if you were asked to solve for the series for ##\cos(u)##? Then let ##u = z^{-1}##?
Ok, I was thinking of doing that. I just dunno, it seems to weird. Obviously then the problem is just the simplest Taylor series in the world just about.

Thanks.

Right...the Laurent series allows you to build into the negative exponents which can capture the information near the singularity of 1/z.
So, if the Taylor series of f(u) is ##\sum_{k = 0}^\infty a_k u^k ## then the equivalent Laurent series for f(1/u) is ##\sum_{k = 0}^{\infty} a_k u^{-k} ##

## What is a Laurent Series?

A Laurent Series is a representation of a complex function as an infinite sum of powers of a variable, including both positive and negative powers.

## What is the Laurent Series for Cos(1/z)?

The Laurent Series for Cos(1/z) is given by the infinite series:
n=0 ((-1)n / (n!) * (z^-n)), where z is a complex variable.

## What is the difference between a Laurent Series and a Taylor Series?

A Taylor Series is a special case of a Laurent Series where all the coefficients for the negative powers of the variable are equal to 0. A Laurent Series includes both positive and negative powers, allowing for a wider range of functions to be represented.

## What is the region of convergence for the Laurent Series of Cos(1/z)?

The region of convergence for the Laurent Series of Cos(1/z) is the annulus 0 < |z| < ∞, which means it converges for all complex numbers except for z = 0.

## How is the Laurent Series of Cos(1/z) used in mathematics?

The Laurent Series of Cos(1/z) is used in complex analysis to study the behavior of functions in the complex plane. It can also be used to approximate values of the function for certain values of z within the region of convergence.

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