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Thread: Complex Analysis - Residues at poles

  1. #1
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    Complex Analysis - Residues at poles

    Let $\displaystyle f(z)=(1+z^2)/7z^3+50iz^2-7z$

    (i). Determine the residue of $\displaystyle f$ at each of its poles
    (ii). Deduce

    $\displaystyle \int_0^{2 pi }(cost/25+7sint)dt=0$
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  2. #2
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    Quote Originally Posted by moolimanj View Post
    Let $\displaystyle f(z)=(1+z^2)/7z^3+50iz^2-7z$
    What makes me angry is that you are in an advanced math class, and still did not learn how to use paranthesis

    We have,
    $\displaystyle f(z) = \frac{1+z^2}{7z^3+60iz^2-7z}$

    The poles arise (possibly) when:
    $\displaystyle 7z^3+60iz^2 - 7z=0$

    Now I am not going to do this problem because the solutions to that equation are messy, are you sure this is the problem?
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    Firstly, about the parenthesis - its my first time using the inbuilt equation editor (which I have never used before) so as long as it looked ok, i was happy rather than spending ages figuring it out. please accept my apologies. I'll have a play around with it to so that i can get it right next time.


    Its definitely the problem. Like you say, its quite messy but rest assured it is whats required since this question gets a total of 8% of the paper.
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    I am sorry, it is not messy, I first read it as a 60 instead of a 50 which made it messy.

    Let me continue the poles arise possibly when:
    $\displaystyle 7z^3+50iz^2-7z=0$
    That is,
    $\displaystyle z(7z^2+50iz-7)=0$
    Thus,
    $\displaystyle z=0 \mbox{ or }z=\pm 48i$.

    It is in the special form $\displaystyle g(z)/h(z)$ where $\displaystyle g,h$ are holomorphic and $\displaystyle h'(c)\not = 0$.

    So the residues can be computed as:
    $\displaystyle \frac{1+z^2}{21z^2+100iz - 7}$ substitute for $\displaystyle z=0,\pm 48i$
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  5. #5
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    Quote Originally Posted by moolimanj View Post
    (ii). Deduce

    $\displaystyle \int_0^{2 pi }(cost/25+7sint)dt=0$
    $\displaystyle \int_0^{2\pi} \frac{\cos t}{25+7\sin t} dt = 0$

    Consider the integral,
    $\displaystyle \oint_{|z|=1} \frac{\frac{1}{2}\left( z + z^{-1} \right)}{25 + \frac{7}{2i}\left( z - z^{-1} \right)}\cdot \frac{1}{zi} \ dz$

    Because if we integrate this using the parametrization $\displaystyle z=e^{it}$ on $\displaystyle 0\leq t\leq 2\pi$ we get percisely the integral you have above.

    Simpify the integral,
    $\displaystyle -\frac{1}{4}\oint_{|z|=1}\frac{z^2+1}{7z^3+50iz-7z} \ dz$

    Above we found that the poles where $\displaystyle z=0,\pm 48i$. Since we are working on the unit circle $\displaystyle |z|=1$ we see that $\displaystyle \pm 48i$ lie outside this circle. Thus, $\displaystyle z=0$ is the only pole.

    So the integral value is,
    $\displaystyle 2\pi i \left( \frac{1}{-4} \right) \cdot \mbox{res}(f,0) = 0 + i \mbox{something}$

    If we equal real and imaginaries we find that the real part is 0. Which is the value of the integral.
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