# Residue question

1. Nov 8, 2011

### Applejacks

1. The problem statement, all variables and given/known data

http://imageshack.us/photo/my-images/593/aaayl.png/

http://imageshack.us/photo/my-images/593/aaayl.png/

2. Relevant equations

3. The attempt at a solution

I already solved part b but I put it up there for part c. I know that there is a singularity at a=cost. However, maximum of cost=1 and a>1 so there is no singularity? Hence the integral equals zero. However, I don't get what the solution from part b tell us

2. Nov 8, 2011

### Dick

The integral of 1*dt for t=0 to 2*pi isn't zero. And 1 doesn't have any singularities either. Because that's a REAL integral, it's not a complex integral. Why don't you try to express it as an integral dz instead of dt where z=e^(it) before you decide it's zero? You should find it has a lot to do with part b.

Last edited: Nov 8, 2011
3. Nov 8, 2011

### Applejacks

$\int\frac{dz}{iz (a-cos(-iln(z)))}$ is what I get. Then using Cauchy's Integral Formula:

2$\pi$i*f(0) should equal to the integral. However, f(0) is undefined because of the log.

4. Nov 8, 2011

### Dick

Use cos(t)=(exp(it)+exp(-it))/2=(z+1/z)/2. Use Euler's formula. Don't bring log's into it. That's big trouble.

5. Nov 9, 2011

### Applejacks

I see. I end up getting the same function in part b but with a -2/i in front.

So we have 2ipi*(sum of res)*-i/2= the integral
However, I'm getting zero for the sum of the residue.
$\frac{1}{(z-(a+\sqrt{a^2-1}))(z-(a-\sqrt{a^2-1})}$

The residues are $\frac{1}{2\sqrt{a^2-1}}$ and-$\frac{1}{2\sqrt{a^2-1}}$

Is this correct?

6. Nov 9, 2011

### Dick

It would be, except you only sum over the residues inside the unit circle. Only one is inside. Which one is it??

7. Nov 9, 2011

### Applejacks

Well I'm looking at the two singularities:

a+$\sqrt{a^2-1}$
a-$\sqrt{a^2-1}$

For a>1, the second one tends to zero. So we should have
2ipi*-i/2 *sum (res)=pi* $\frac{-1}{2\sqrt{a^2-1}}$?

8. Nov 10, 2011

### Dick

The real integral you started with certainly isn't negative. Check that again. I don't think you did it very carefully.

9. Nov 11, 2011

### Applejacks

Okay I found my mistake. It should have been
$\frac{2z}{(z-(a+\sqrt{a^2-1}))(z-(a-\sqrt{a^2-1})}$

Now the residues are:
$\frac{-a-\sqrt{a^2-1}}{\sqrt{a^2-1}}$
$\frac{a-\sqrt{a^2-1}}{\sqrt{a^2-1}}$

So finally I have 2ipi*sum(res)=2ipi* -2=-4ipi

Is this right?

10. Nov 11, 2011

### Dick

NO. i) Look at the integral you are trying to solve. It's positive and real, right? -4ipi is neither. And ii) part b gave you the residue you need and you said you had already solved that part. It looked like you had it right in post 5. Why don't you start this whole thing from the beginning and show all of your steps and I'm sure someone will point out where you made a mistake.

11. Nov 11, 2011

### Applejacks

Sorry ignore my last post. I forgot to transform dt into dz.

Here is my working out:
http://imageshack.us/f/215/unled2lwl.png/

I just don't understand why I have to use the positive residue. What piece of information tells me that the negative part isn't inside the unit circle.

12. Nov 11, 2011

### Dick

The poles of your function are at z=a+sqrt(a^2-1) and z=a-sqrt(a^2-1). Pick which one satisfies |z|<1. It isn't the first one, is it? Why not??

13. Nov 11, 2011

### Applejacks

Well that's sort of my point. The first one is outside the unit circle since a>1, hence z becomes >1. Only the second one is inside the circle so how come I'm not using that one instead?

14. Nov 11, 2011

### Dick

You are! Call the two poles r+ and r-. Evaluate the residue at r-. Your function is proportional to 1/((z-r+)(z-r-)). So the residue is lim z->r- (z-r-)*1/((z-r+)(z-r-))=lim z->r- 1/(z-r+)=1/((r-)-(r+)). That's the one one that gives you the negative residue. Work it out carefully!

Last edited: Nov 11, 2011
15. Nov 11, 2011

### Applejacks

I see now. However, my final answer should have 2pi instead of pi/2 right? I reworked it and that's what I got.

16. Nov 12, 2011

### Dick

Yes, it should be 2*pi/sqrt(a^2-1).