# Help me with the residue.

1. Jun 20, 2013

### aiisshsaak

Hello guys,

2. Jun 20, 2013

### HallsofIvy

Staff Emeritus
"Confirm"? You mean you have already found an answer? Great! Tell us what you got and we will try to confirm it for you.

3. Jun 20, 2013

### aiisshsaak

Right :p

4. Jun 20, 2013

### Office_Shredder

Staff Emeritus
The answer should be a number, not a function.

Also, there are two different points where you might be interested in the residue - at z=0 and z=1. Which one are you supposed to find?

5. Jun 20, 2013

### aiisshsaak

at z°=1

here is how I've done it ..

Res=limit [f(z)*(z°-z)] as z goes to z°
=limit [e1/z/[STRIKE](1-z)[/STRIKE]*[STRIKE](z°-z)[/STRIKE]] as z goes to z°
=limit [e1/z] as z goes to z° ''which equals to 1''
and finally I plugged in z=z°=1 and I got: Res=e1/1=e≈2.72

am I wrong somewhere?

6. Jun 20, 2013

### Office_Shredder

Staff Emeritus
That looks correct except that it should be f(z)*(z-z0) which means you're off by a minus sign.

Usually "exp" refers to the function ex, not the number e itself, which is why I wrote the first part of my previous post.

7. Jun 20, 2013

### aiisshsaak

Thx guys for these replies..I just want to know one last thing. using the residue theorem, what is the integral of f(z) ...at |z|=1/2

Thx again :)

8. Jun 20, 2013

### Office_Shredder

Staff Emeritus
Your question isn't very well posed... Are you integrating around the circle defined by |z|=1/2?

9. Jun 21, 2013

### aiisshsaak

Im sry about that, yes integration around the circle .

10. Jun 23, 2013

### aiisshsaak

anyone can help me with the integration plz? f(z)=e1/z/(1-z)
∫f(z)dz around the circle |z|=1/2 ?

11. Jun 23, 2013

### daveyrocket

You need to identify the poles within the circle |z| = 1/2. You might consider expanding the exponential in a series.

12. Jun 23, 2013

### aiisshsaak

All I need is the result, so can u provide me with that? thx

13. Jun 23, 2013

### Office_Shredder

Staff Emeritus
No, we can't. We can help guide your work but we will not tell you what the answer is.

There is a single pole at z=0 inside of the circle (it should be fairly obvious from looking at the function). Can you find the residue at that pole?

14. Jun 23, 2013

### aiisshsaak

But I thought the pole is z=1, no?

15. Jun 24, 2013

### daveyrocket

This function has more than one pole. Look at the 1/z in the exponential.

16. Jun 24, 2013

### aiisshsaak

Right, ty person ;)