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Math Help - Complex conjugate?

  1. #1
    Super Member Showcase_22's Avatar
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    Complex conjugate?

    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?
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  2. #2
    Super Member Showcase_22's Avatar
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    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?
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  3. #3
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    Quote Originally Posted by Showcase_22 View Post
    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
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    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
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  5. #5
    Super Member Showcase_22's Avatar
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    <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?
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  6. #6
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    Yes, use Cauchy's Integral formula. Then in that case, just take derivatives even thought that's still implicitly residues.
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  7. #7
    Super Member Showcase_22's Avatar
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    Ah, I think I get it!
    Last edited by Showcase_22; November 29th 2009 at 12:13 PM.
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  8. #8
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    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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