1. ## Laurent Series

Hey guys, I'm having a bit of trouble with the creating Laurent series representations of functions. Specifically, I have no idea what the domain does to the series representation.

Anyways, this is what I'm working on...

$\frac{e^z}{(z+1)^2}$ for $0 < |z+1| < \infty$

This is where I go with this?...

$\frac{e^z}{(z+1)^2}= \frac{1}{(z+2)^2} \displaystyle\sum_{n=0}^{\infty}\frac{(z+1)^n}{(n+ 1)!}$

No idea where to go from here.

Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

Thanks for the help in advance.

2. Use the substitution $u=z+1$ , then:

$f(z)=\dfrac{e^{u-1}}{u^2}=\dfrac{1}{e}\displaystyle\sum_{n\geq 0}\dfrac{u^{n-2}}{n!}=\ldots(0<|u|<+\infty)$

Originally Posted by bakerconspiracy
Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?

Yes, it is.

3. Originally Posted by bakerconspiracy
Hey guys, I'm having a bit of trouble with the creating Laurent series representations of functions. Specifically, I have no idea what the domain does to the series representation.

Anyways, this is what I'm working on...

$\frac{e^z}{(z+1)^2}$ for $0 < |z+1| < \infty$

This is where I go with this?...

$\frac{e^z}{(z+1)^2}= \frac{1}{(z+2)^2} \displaystyle\sum_{n=0}^{\infty}\frac{(z+1)^n}{(n+ 1)!}$

No idea where to go from here.
Write the function as $\displaystyle \frac{e^z}{(z+1)^2}= \frac{e^{-1}e^{z+1}}{(z+1)^2}= e^{-1}\sum_{n=0}^\infty\dfrac{(z+1)^{n-2}}{n!}$. That is the Laurent expansion in powers of z+1.

Originally Posted by bakerconspiracy
Also, if a function is analytic everywhere in a domain, is the Laurent series representation just the Taylor series?
Yes, provided that you use the expansion about the same point in both cases.

Edit. Sorry, didn't notice that FernandoRevilla got there first.

4. Thank you both. I'm starting to understand this better now.