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Thread: Rouché's theorem

  1. #1
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    Rouché's theorem

    I have a question that is as follows,

    Let $\displaystyle p(z)=z^{2}-ce^{-z}$ where $\displaystyle 0<c<1$

    Use Rouches Theorem to show that if $\displaystyle c<{\frac{1}{e}}$ then there are two zeros of p(z) within the unit circle.

    Now i chose $\displaystyle g(z)=z^{2}$ and $\displaystyle h(z)=-ce^{-z}$

    and then evaluated these at z=1, as $\displaystyle \left | z \right |=1$ on the boundary of the circle to get,

    $\displaystyle \left | h(1) \right |=\left | \frac{-c}{e} \right |=\frac{c}{e}$

    $\displaystyle \left | g(1) \right |=\left | 1^{2} \right |=1$

    i then said that by rouches theorem, $\displaystyle g(z)+h(z)$ (i.e p(z)) has the same number of zeros as g(z), i.e. 2, if

    $\displaystyle \left | h(1) \right |<\left | g(1) \right |$

    which leads to me concluding that this is when

    $\displaystyle c<e$

    Does anyone know what i am doing wrong?
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  2. #2
    Senior Member Tinyboss's Avatar
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    Looks right to me. Of course, you have also proved the problem as stated. ;-) But I think you're right, there's probably an error in the statement.
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  3. #3
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    Yes, i guess it does satisfy it. But the next part of the question is that if c>1/e then we might think that there is only one solution in the cicle, then by setting c=1/2, show using the integral form of the arguement principle that there is only one zero in this case.
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