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Math Help - Complex roots

  1. #1
    Super Member Deadstar's Avatar
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    Complex roots

    This should be the easiest part of my module but i just cant figure it out...

    Let f(z) = \frac{e^{2 i \pi z}}{1 + z^3}. Find the singularities and classify them.

    So need to find z when z^3 = -1.
    One root is -1. The others we get from DeMoivres Thm...
    I seem to think it should be something like z = \cos(\frac{k \pi}{3}) + i\sin(\frac{k \pi}{3}) for k = 1,2,3 but that gets me i \frac{\sqrt(3)}{2} \pm \frac{1}{2} instead of \frac{1}{2} \pm i \frac{\sqrt(3)}{2}...

    Also is there an easy way to classify the singularities? I know they are simple poles but im not sure how to prove it...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    This should be the easiest part of my module but i just cant figure it out...

    Let f(z) = \frac{e^{2 i \pi z}}{1 + z^3}. Find the singularities and classify them.

    So need to find z when z^3 = -1.
    One root is -1. The others we get from DeMoivres Thm...
    I seem to think it should be something like z = \cos(\frac{k \pi}{3}) + i\sin(\frac{k \pi}{3}) for k = 1,2,3 but that gets me i \frac{\sqrt(3)}{2} \pm \frac{1}{2} instead of \frac{1}{2} \pm i \frac{\sqrt(3)}{2}...
    The roots are of the form cos(\frac{(\pi+ k 2\pi)i}{3})+ i sin(\frac{(\pi+ k 2\pi)i}{3}. You start with the \pi i of -1 and add multiples of 2\pi, not \pi, and divide by 3.

    Also is there an easy way to classify the singularities? I know they are simple poles but im not sure how to prove it...
    You can factor the function into \frac{e^{2i\pi z}}{(z-z_0)(z-z_1)(z-z_2)} where z_0, z_1, z_2 are the zeros of the denominator. Now you can see that, for example, in a neighborhood of z_0, you have a function analytic in that neighborhood ( \frac{e^{2i\pi z}}{(z-z_1)(z-z_2)} times \frac{1}{z-z_0} which clearly has a pole of order 1 at z_0- write the first as a Taylor's series about z-z_0 and divide by z-z_0 getting as Laurent series having -1 as the only negative power.
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