# Help With Laurent Series

1. Nov 9, 2004

### KeithF40

I need help with a problem from Complex Analysis. The directions say find the Laurent series that converges for 0<|z|<R and determine the precise region of convergence. The expression is : e^z/(z-z^2). I understand how to do the other 7 problems in this section but not this one. Can someone please help me with this one as Ive spent hours trying to figure it out.

2. Nov 9, 2004

### ReyChiquito

$$f(z)=\frac{e^z}{z-z^2}=\frac{e^z}{z(1-z)}$$

remember that

$$e^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$

where is its radio of convergence?

now

for $|z|<1$

$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$

and for $1<|z|$

$$\frac{1}{1-z}=-\frac{1}{z(1-\frac{1}{z})}=-\frac{1}{z}\sum_{n=0}^{\infty}\frac{1}{z^n}$$

can you take it from here?

Last edited: Nov 9, 2004
3. Nov 9, 2004

### KeithF40

Thanks for the info so far but not really. The other questions seemed to much easier to do than this one. Im having trouble combining the different series up together. Thanks if you can help me more with this.

4. Nov 10, 2004

### KeithF40

I got this solution :
(1/z)*(E(z^n/n!))*(E(z^n)) with both summations from n=0 to inf. The radius of convergence that I found for this was to be 0<|z|<1. Is this correct and if so is there anyway to clean this up and express it as one summation.

5. Nov 10, 2004

### ReyChiquito

hmmmmm.... ill do one and youll do the other ok?

for $|z|<1$ youll have

$$f(z)=\frac{e^z}{z(1-z)}$$

$$f(z)=\frac{1}{z}\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=0}^{\infty}z^n\right)$$

and using (can you proove this?, i have)

$$\left(\sum_{n=0}^{\infty}a_{n}z^n\right)\left(\sum_{n=0}^{\infty}b_{n}z^n\right)=\sum_{n=0}^{\infty}c_{n}z^n$$

where

$$c_{n}=\sum_{k=0}^{n}a_{k}b_{n-k}$$

implies that

$$f(z)=\frac{e^z}{z(1-z)}=\frac{1}{z}\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=0}^{\infty}z^n\right)=\frac{1}{z}\sum_{n=0}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}\right]z^n$$

so, around $|z|<1$ the Laurent expantion will be

$$\frac{e^z}{z(1-z)}=\frac{1}{z}+\sum_{n=1}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}\right]z^{n-1}=\frac{1}{z}+\sum_{n=0}^{\infty}\left[\sum_{k=0}^{n+1}\frac{1}{k!}\right]z^{n}$$

there is my part of the deal... now, can you take it from there?

EDIT. I was answering 3 when you posted 4 :tongue:

Last edited: Nov 10, 2004
6. Nov 10, 2004

### KeithF40

So is my answer right or wrong. Im not sure what you said about it also I dont know if its supposed to be only one summation and if so I dont know how to make it one summation. I had the nested summations like you did before but I dont know if that is the right final answer. Thanks.

7. Nov 11, 2004

### ReyChiquito

your answer is correct, but is incomplete, as you need now to do the rest of the series (when |z|>1). Besides, if you leave the series expressed as a product, is not clear that it has taylor/laurent form, but is correct.

Remember, these expantion valid for z around 0.

8. Nov 11, 2004

### ReyChiquito

oh, and

$$\left(\sum_{n=0}^{\infty}a_{n}z^n\right)\left(\sum_{n=0}^{\infty}\frac{b_{n}}{z^n}\right)=\sum_{n=0}^{\infty}c_{n}{z^n}+\sum_{n=1}^{\infty}\frac{d_{n}}{z^n}$$

where

$$c_{n}=\sum_{k=0}^{\infty}a_{n+k}b_{k}$$

and

$$d_{n}=\sum_{k=0}^{\infty}a_{k}b_{n+k}$$

Last edited: Nov 12, 2004