Hello! I have to evaluate:

I decided to start by getting rid of the imaginary part in the denominator. If then .Quote:

HOWEVER! . Unfortunately, this hasn't removed the imaginary ppart from the denominator.

What's going wrong?

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- Nov 29th 2009, 09:31 AMShowcase_22Complex conjugate?
Hello! I have to evaluate:

Quote:

HOWEVER! . Unfortunately, this hasn't removed the imaginary ppart from the denominator.

What's going wrong? - Nov 29th 2009, 09:48 AMShowcase_22
AhhH! sorry, of course it's real! .

Anyway, i've now run into a new problem. I can get the integral into the form:

.

What's the best way to tackle this integral? - Nov 29th 2009, 09:51 AMtonio
- Nov 29th 2009, 10:46 AMshawsend
- Nov 29th 2009, 11:13 AMShowcase_22Quote:

I'm also afraid that I haven't learn what residues are yet and I don't appear to have anything in my notes about it.

Is there a way of doing this question without using residues? - Nov 29th 2009, 11:19 AMshawsend
Yes, use Cauchy's Integral formula. Then in that case, just take derivatives even thought that's still implicitly residues.

- Nov 29th 2009, 11:38 AMShowcase_22
Ah, I think I get it!

- Nov 29th 2009, 11:43 AMshawsend
Yes. Try and get in the habit of checking it with Mathematica if you have access to it:

Code:`In[115]:=`

NIntegrate[(Exp[z^2]/(z - 1)^3)*2*I*

Exp[I*t] /. z -> 2*Exp[I*t],

{t, 0, 2*Pi}]

N[6*E*Pi*I]

Out[115]=

1.2434497875801753*^-14 +

51.23840533603834*I

Out[116]=

0. + 51.2384053360414*I