Complex conjugate?

**Showcase_22** · November 29th 2009, 09:31 AM

Hello! I have to evaluate:

$\int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}~dt$

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

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

What's going wrong?

**Showcase_22** · November 29th 2009, 09:48 AM

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

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

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

What's the best way to tackle this integral?

**tonio** · November 29th 2009, 09:51 AM

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 $w=(2e^{it}-1)^3$ then $\overline{w}=(2e^{-it}-1)^3$ .

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

What's going wrong?

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

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

Tonio

**shawsend** · November 29th 2009, 10:46 AM

Originally Posted by Showcase_22

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

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

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

What's the best way to tackle this integral?

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

$\int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}dt=\mathop\oint\limits_{|z|=2} \frac{e^{z^2}}{(z-1)^3} dz$

**Showcase_22** · November 29th 2009, 11:13 AM

$<br /> <br /> \int_0^{2 \pi} \frac{e^{4e^{2it}}(2ie^{it})}{(2e^{it}-1)^3}dt=\mathop\oint\limits_{|z|=2} \frac{e^{z^2}}{(z-1)^3} dz<br />$

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?

**shawsend** · November 29th 2009, 11:19 AM

Yes, use Cauchy's Integral formula. Then in that case, just take derivatives even thought that's still implicitly residues.

**Showcase_22** · November 29th 2009, 11:38 AM

Ah, I think I get it!

**shawsend** · November 29th 2009, 11:43 AM

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

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