Laurent Series/ Laurent Series Expansion

**FatherMike** · October 5th 2010, 06:12 PM

Hello everyone,

I just had a couple questions about Laurent series.

How do I find the Laurent series of:
$z^{2}cos(\frac{1}{3z})$ in the region $\left | z \right | > 0$

How do I find the Laurent series expansion of $(z^{2} - 1)^{-2}$ valid in the regions:
a) $0 < \left | z-1 \right |< 2$

b) $\left | z+1 \right |> 2$

Any help would be great. Thanks for your time!

**TheEmptySet** · October 5th 2010, 08:41 PM

Originally Posted by FatherMike

Hello everyone,

I just had a couple questions about Laurent series.

How do I find the Laurent series of:
$z^{2}cos(\frac{1}{3z})$ in the region $\left | z \right | > 0$

How do I find the Laurent series expansion of $(z^{2} - 1)^{-2}$ valid in the regions:
a) $0 < \left | z-1 \right |< 2$

b) $\left | z+1 \right |> 2$

Any help would be great. Thanks for your time!

For the first one just use the Taylor series for cosine centered at zero.

$\displaystyle g(z)=\cos(z)=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n)!}$

Then your series is

$\displaystyle z^{2}cos\left(\frac{1}{3z}\right)=z^2g\left( \frac{1}{3z}\right)$

For the 2nd part this will get you started

First note that

$\displaystyle \frac{1}{(1-z^2)^2}=\frac{1}{(1-z)^2(1+z)^2}$

lets focus on the $0 < |z-1|<2$

let $\displaystyle h(z)=\frac{1}{1+z}=\frac{1}{2+(z-1)}=\frac{1}{2}\left(\frac{1}{1+\frac{z-1}{2}} \right) =\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-1)^n$

Then

$\displaystyle h'(z)=\frac{-1}{(1+z)^2}=\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^{n+1}}(z-1)^{n-1} = \sum_{n=0}^{\infty}\frac{(n+1)(-1)^{n+1}}{2^{n+2}}(z-1)^n$

Can you finish from here?

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