Get Expert Help with Laurent Series for Convergence of e^z/(z-z^2)"

[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 I've spent hours trying to figure it out.

 

[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?

 

[KeithF40]

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

 

[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.

 

[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 expansion 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

 

[KeithF40]

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

 

[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 expansion valid for z around 0.

 

[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

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