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Thread: Solving integral using residue theorem

  1. #1
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    Solving integral using residue theorem

    I have the following integral:

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

    since:

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

    I rewrite the integral and get the following:

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

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

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

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

    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 $\displaystyle z$ I get the following:

    $\displaystyle \displaystyle{z_{1}=2$

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

    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 $\displaystyle \int_{|z|=1}\ldots dz$ instead of $\displaystyle \int_0^{2\pi}\ldots dz$. Now find the residue for the pole $\displaystyle z_2,\;\;(|z_2|<1)$

    Regards.

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

    Regards.

    ---
    Fernando Revilla

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

    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 $\displaystyle x_2=-1/2$ is:


    $\displaystyle \begin{array}{rcl}
    \textrm{Res}[f,-1/2]=
    \\ \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)'=

    \\ -\dfrac{1}{4i}\displaystyle\lim_{z \to -1/2}{\left(\dfrac{z}{(z-2)^2}\right)'=

    \\\ldots\;=

    \\ -\dfrac{3}{125i}
    \end{array}
    $

    Then,

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

    Regards.

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

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


    How does the following:

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

    becomes:

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

    Thank you for your response.

    Kind regards
    Last edited by 4Math; Nov 12th 2010 at 01:16 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:

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

    becomes:

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

    Using two theorems:

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

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

    then:

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


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

    $\displaystyle \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})}
    $

    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:

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

    to the following:

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

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

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

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

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

    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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