# Homework Help: Complex analysis

1. Mar 8, 2009

### asi123

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

Hey guys.
I have this integral, I tried to use trigo, tried to use the complex expression but nothing worked, can I please have some help?

Thanks a lot.

2. Relevant equations

3. The attempt at a solution

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2. Mar 8, 2009

### gabbagabbahey

The complex Residues method should work fine; why don't you show us what you tried for that method...

3. Mar 9, 2009

### asi123

Ok, so I tried to use the residues theorem.
I did the substitute but ended up with that really ugly expression, should I use the residues theorem on it?

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4. Mar 9, 2009

5. Mar 9, 2009

### asi123

http://img8.imageshack.us/img8/1059/scan0015p.jpg [Broken]

Thanks a lot.

Last edited by a moderator: May 4, 2017
6. Mar 9, 2009

### gabbagabbahey

Okay, so far so good....now find the 4 roots of the denominator in order to find the poles....which of those poles lies within your contour?

7. Mar 9, 2009

### gabbagabbahey

To make your calculations easier, you should note that if $|z^2|>1$ then so is $|z|$; and $3+2\sqrt{2}>1$.

Also note that $$\sqrt{3-2\sqrt{2}}=\sqrt{2}-1$$

8. Mar 9, 2009

### asi123

Oh, I've only now noticed your calculations tips

Anyway, this is what I did:

http://img27.imageshack.us/img27/4732/scan0016y.jpg [Broken]

I've to warn you, The numbers are awful.

Does it seems right?

Thanks a lot.

Last edited by a moderator: May 4, 2017
9. Mar 9, 2009

### gabbagabbahey

They sure are!

Luckily, they're also correct....you can simplify them though...after a little algebra you should find that $A=B=\frac{-1}{\sqrt{2}}$ and so your final result becomes $2\pi(\sqrt{2}-1)$

10. Mar 9, 2009

Thanks a lot