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Math Help - Complex Analysis - Residues at poles

  1. #1
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    Complex Analysis - Residues at poles

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

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

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

    The poles arise (possibly) when:
    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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  4. #4
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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:
    7z^3+50iz^2-7z=0
    That is,
    z(7z^2+50iz-7)=0
    Thus,
    z=0 \mbox{ or }z=\pm 48i.

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

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

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

    Consider the integral,
    \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 z=e^{it} on 0\leq t\leq 2\pi we get percisely the integral you have above.

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

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

    So the integral value is,
    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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