# Singularity points + Laurent series

1. Mar 27, 2014

### skrat

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

2. Mar 27, 2014

### vela

Staff Emeritus
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.

3. Mar 27, 2014

### skrat

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

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.