# Complex conjugate?

• Nov 29th 2009, 09:31 AM
Showcase_22
Complex conjugate?
Hello! I have to evaluate:

Quote:

$\displaystyle \int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}~dt$
I decided to start by getting rid of the imaginary part in the denominator. If $\displaystyle w=(2e^{it}-1)^3$ then $\displaystyle \overline{w}=(2e^{-it}-1)^3$.

HOWEVER! $\displaystyle w \overline{w}=(2e^{it}-1)^3(2e^{-it}-1)^3=(4-2e^{-it}-2e^{it}+1)^3$. Unfortunately, this hasn't removed the imaginary ppart from the denominator.

What's going wrong?
• Nov 29th 2009, 09:48 AM
Showcase_22
AhhH! sorry, of course it's real! $\displaystyle w \overline{w}=5-4 \cos t$.

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

$\displaystyle \oint_0^{2 \pi} \frac{(2ie^{4e^{2it}+it})(2e^{it}-1)^3}{5-4\cos t}~dt$.

What's the best way to tackle this integral?
• Nov 29th 2009, 09:51 AM
tonio
Quote:

Originally Posted by Showcase_22
Hello! I have to evaluate:

I decided to start by getting rid of the imaginary part in the denominator. If $\displaystyle w=(2e^{it}-1)^3$ then $\displaystyle \overline{w}=(2e^{-it}-1)^3$.

HOWEVER! $\displaystyle w \overline{w}=(2e^{it}-1)^3(2e^{-it}-1)^3=(4-2e^{-it}-2e^{it}+1)^3$. Unfortunately, this hasn't removed the imaginary ppart from the denominator.

What's going wrong?

Of course it has: for $\displaystyle z\in\mathbb{C}\,,\,z\overline z=|z|^2\in\mathbb{R}^+$.

You're probably getting confused with $\displaystyle -2e^{it}-2e^{-it}=-4\,Re(e^{it})=-4\cos t\in\mathbb{R}$

Tonio
• Nov 29th 2009, 10:46 AM
shawsend
Quote:

Originally Posted by Showcase_22
AhhH! sorry, of course it's real! $\displaystyle w \overline{w}=5-4 \cos t$.

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

$\displaystyle \oint_0^{2 \pi} \frac{(2ie^{4e^{2it}+it})(2e^{it}-1)^3}{5-4\cos t}~dt$.

What's the best way to tackle this integral?

Don't mix the closed-contour notation with the end-point notation. Just let $\displaystyle z=2e^{it}$ and switch the original integral to a contour integral that can be evaluated via residue integration:

$\displaystyle \int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}dt=\mathop\oint\limits_{|z|=2} \frac{e^{z^2}}{(z-1)^3} dz$
• Nov 29th 2009, 11:13 AM
Showcase_22
Quote:

$\displaystyle \int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}dt=\mathop\oint\limits_{|z|=2} \frac{e^{z^2}}{(z-1)^3} dz$
Ironically I started with the RHS and ended up with the LHS!

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 AM
shawsend
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 AM
Showcase_22
Ah, I think I get it!
• Nov 29th 2009, 11:43 AM
shawsend
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