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Math Help - Rouché's theorem

  1. #1
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    Rouché's theorem

    I have a question that is as follows,

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

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

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

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

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

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

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

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

    which leads to me concluding that this is when

    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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