# Singularity points + Laurent series

• skrat
In summary, the singularity point for f(z) = (sin(3z)-3z)/z^5 is at z=0, which is a pole of second order. The Laurent series for this function simplifies to -3^3/3!*z^-2 + 3^5/5! + 3^7/7!*z^2 + ... The regular part is 3^5/5! + 3^7/7!*z^2 + ..., while the principal part is -3^3/3!*z^-2.
skrat

## Homework Statement

Find and determine the type of singularity points for ##f(z)=\frac{\sin(3z)-3z}{z^5}##. Also calculate the regular and main part of Laurent series around those points.

## The Attempt at a Solution

I am already having troubles with the first part.

Singularity point is of course ##z=0##. For ##z=0## I can use Taylor expansion for ##\sin## which than gives me

##f(z)=\frac{\sin(3z)-3z}{z^5}\doteq \frac{(3z-\frac{(3z)^3}{3!}+\frac{(3z)^5}{5!}-\cdots )-3z}{z^5}=\frac{z^3(-\frac{3^3}{3!}+\frac{(3^5z^2}{5!}-\cdots )}{z^5}=\frac{-\frac{3^3}{3!}+\frac{3^5z^2}{5!}-\cdots }{z^2}##

Now I have no idea how to determine the type of singularity point. However, if I would have to guess I would say that in ##z=0## is pole of second order.

If that is ok, I would be happy to get a hint on the second part of this problem - Laurent series.

That's right. It's a pole of order 2, and you've essentially derived the Laurent series. Just simplify what you have to
$$f(z) = -\frac{3^3}{3!}\frac{1}{z^2} + \frac{3^5}{5!} - \frac{3^7}{7!} z^2 + \cdots.$$ Now you just have to identify the regular and principal parts of the series.

##f(z)=-\frac{3^3}{3!}\frac{1}{z^2}+\frac{3^5}{5!}-\frac{3^7}{7!}z^2+\frac{3^9}{9!}z^4-\cdots ##

Where principal part is ##-\frac{3^3}{3!}\frac{1}{z^2}## and regular part is ##\frac{3^5}{5!}-\frac{3^7}{7!}z^2+\frac{3^9}{9!}z^4- \cdots ##

And this should be it.

## 1. What is a singularity point?

A singularity point is a point on a complex plane where a function becomes infinite or undefined. It is often denoted as z0 and can be either a finite or an infinite value.

## 2. What is a Laurent series?

A Laurent series is a type of complex power series that represents a function as an infinite sum of terms with both positive and negative powers of the variable. It is useful for representing functions that have singularities, as it can converge in regions outside of the singularity point.

## 3. How do you find the Laurent series of a function?

To find the Laurent series of a function, we use the process of partial fraction decomposition to rewrite the function as a sum of simpler terms. Then, we use the known Taylor series expansions for these simpler terms to write the Laurent series.

## 4. What is the difference between a Taylor series and a Laurent series?

The main difference between a Taylor series and a Laurent series is that a Taylor series represents a function as an infinite sum of only positive powers of the variable, while a Laurent series includes both positive and negative powers. This allows the Laurent series to represent functions with singularities, while the Taylor series can only approximate smooth functions.

## 5. How are Laurent series used in complex analysis?

Laurent series are used in complex analysis to study the behavior of functions at singularities. They can also be used to evaluate integrals and solve differential equations in the complex plane. Additionally, Laurent series are used in the study of analytic continuation, which involves extending a function beyond its domain of convergence.

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