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Math Help - Solving integral using residue theorem

  1. #1
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    Solving integral using residue theorem

    I have the following integral:

    \displaystyle\int_{0}^{2\pi}\frac{1}{(4sinx+3i)^{2  }}= \left [{\begin{array}{ll}z=e^{ix} \,\,\\\frac{dz}{iz}=dx\end{array}\right]=

    since:

    \displaystyle sin x=\frac{1}{2i}(z-z^{-1})

    I rewrite the integral and get the following:

    \displaystyle=\int_{0}^{2\pi}\frac{1}{((\frac{2z^2-2}{iz})+3i)^{2}}\frac{dz}{iz}=

    \displaystyle=\int_{0}^{2\pi}\frac{1}{(\frac{2z^2-2-3z}{iz})^{2}}\frac{dz}{iz}=

    \displaystyle=\int_{0}^{2\pi}\frac{-z^{2}}{(2z^2-2-3z)^{2}}\frac{dz}{iz}=

    \displaystyle=\int_{0}^{2\pi}\frac{-z}{(2z^2-2-3z)^{2}}\frac{dz}{i}

    How do I get rid off the minus in the numerator in the last integral? Is this correct so far? I need some help and suggestions to continue the solution.

    When I solve for z I get the following:

    \displaystyle{z_{1}=2

    \displaystyle{z_{2}=-\frac{1}{2}

    I don't know if that is correct.

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    You should write \int_{|z|=1}\ldots dz instead of \int_0^{2\pi}\ldots dz. Now find the residue for the pole z_2,\;\;(|z_2|<1)

    Regards.

    ---
    Fernando Revilla
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    You should write \int_{|z|=1}\ldots dz instead of \int_0^{2\pi}\ldots dz. Now find the residue for the pole z_2,\;\;(|z_2|<1)

    Regards.

    ---
    Fernando Revilla

    I am trying to find the residue using the following:
    \displaystyle\mathrm{Res}_{z \to z_2} \frac{f_1(z)}{f_2(z)}=\frac{f_1(z_2)}{f{'}_2(z_2)}

    In other words I just derive the denominator in the last integral in my previous post. But the answer I got is incorrect. Is there any other way finding the residue?

    Kind regards
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    The residue at x_2=-1/2 is:


    \begin{array}{rcl}<br />
\textrm{Res}[f,-1/2]=<br />
\\ \displaystyle\lim_{z \to -1/2}{\dfrac{1}{i}\cdot}\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)'=<br /> <br />
\\ -\dfrac{1}{4i}\displaystyle\lim_{z \to -1/2}{\left(\dfrac{z}{(z-2)^2}\right)'=<br /> <br />
\\\ldots\;=<br /> <br />
\\ -\dfrac{3}{125i}<br />
\end{array} <br />

    Then,

    I=2\pi i\textrm{Res}[f,-1/2]=-\dfrac{6\pi}{125}

    Regards.

    Fernando Revilla
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  5. #5
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    Quote Originally Posted by FernandoRevilla View Post

    \displaystyle \left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)


    How does the following:

    \displaystyle\frac{-z}{(2z^2-2-3z)^{2}}

    becomes:

    \displaystyle\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)<br />

    Thank you for your response.

    Kind regards
    Last edited by 4Math; November 12th 2010 at 02:16 PM.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by 4Math View Post
    How does the following:

    \displaystyle\frac{-z}{(2z^2-2-3z)^{2}}

    becomes:

    \displaystyle\left(\dfrac{-z}{4(z-2)^2(z+(1/2))^2}\cdot (z+(1/2))^2\right)<br />

    Using two theorems:

    (i) If z_1,\ldots,z_n are the roots of the polinomyal of degree n:

    f(z)=a_nz^n+\ldots+a_1z+a_0\in{\mathbb{C}}[z]

    then:

    f(z)=a_n(z-z_1)\cdot\ldots\cdot(z-z_n)


    (ii) If z_0 is a pole of f with multiplicity k, then:

    \textrm{Res}[f,z_0]=\dfrac{1}{(k-1)!}\displaystyle\lim_{z \to z_0}\dfrac{d^{k-1}}{dz^{k-1}}{\left(f(z)(z-z_0\right)^k})}<br />

    Regards.
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  7. #7
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    I follow the two applied theorems that you are refering to.

    But I still can't understand how you factor:

    \displaystyle{(2z^2-2-3z)^{2}

    to the following:

    {4(z-2)^2(z+(1/2))^2}

    Kind regards
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Using (i) :

    (2z^2-2-3z)^{2}=[\;2(z-2)(z+(1/2))\;]^2=4(z-2)^2(z+(1/2))^2

    Regards.
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  9. #9
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    Quote Originally Posted by FernandoRevilla View Post
    Using (i) :

    (2z^2-2-3z)^{2}=[\;2(z-2)(z+(1/2))\;]^2=4(z-2)^2(z+(1/2))^2

    Regards.
    I can now clearly see the equality of these two expressions. I have to blame it that it was late night when I looked at it last time.

    Thank you very much indeed.

    Kind regards
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