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Math Help - Complex Analysis

  1. #1
    Senior Member Dinkydoe's Avatar
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    Complex Analysis

    I want to show that ze^{a-z}-1=0, with  a> 1, has exactly one root in the open disc D(0,1). I'm trying to do this with Rouche's theorem, that is:

    finding a suitable function g(z) with exactly one zero in D(0,1), that satisfies |f(\zeta)-g(\zeta)|<|f(\zeta)|+|g(\zeta)| for all \zeta\in D(0,1)

    I believe it's not all too hard to find such function, like g(z)=ze^{a-z} however showing the desired inequality is for me problematic. Any help/suggestions?
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  2. #2
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    Quote Originally Posted by Dinkydoe View Post
    I want to show that ze^{a-z}-1=0, with  a> 1, has exactly one root in the open disc D(0,1). I'm trying to do this with Rouche's theorem, that is:

    finding a suitable function g(z) with exactly one zero in D(0,1), that satisfies |f(\zeta)-g(\zeta)|<|f(\zeta)|+|g(\zeta)| for all \zeta\in D(0,1)

    I believe it's not all too hard to find such function, like g(z)=ze^{a-z} however showing the desired inequality is for me problematic. Any help/suggestions?
    Try writing it as z = e^{z-a}.
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  3. #3
    Senior Member Dinkydoe's Avatar
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    I tried, with f(z)= z-e^{z-\alpha}. Then with Rouche trying to show that f(z) has only 1 zero in D(0,1)

    however it seems harder then I thought finding a suitable function g(z) with exactly one zero in D(0,1) and such that |f(\zeta)-g(\zeta)|< |f(\zeta)|+|g(\zeta)| for all \zeta in D(0,1)

    It must be a pretty clever choice for g since the most obvious choices like g(z)=z is not going to work.
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Dinkydoe View Post
    I tried, with f(z)= z-e^{z-\alpha}. Then with Rouche trying to show that f(z) has only 1 zero in D(0,1)

    however it seems harder then I thought finding a suitable function g(z) with exactly one zero in D(0,1) and such that |f(\zeta)-g(\zeta)|< |f(\zeta)|+|g(\zeta)| for all \zeta in D(0,1)

    It must be a pretty clever choice for g since the most obvious choices like g(z)=z is not going to work.
    The usual version of Rouché's theorem says that if |f(z)| < |g(z)| on the boundary of some domain G, then g and f + g have the same number of zeros in G. The result you have quoted is the symmetric version of Rouché's theorem, though even in that version the inequality only needs to hold on the boundary of G, not in the whole of G. It is the usual version, not the symmetric version, that is useful here.

    The inequality |-e^{z-a}| < |z| holds whenever z is on the boundary of D(0,1) (that is, when |z| = 1), so Rouché's theorem applies there.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    Ah thank you so much. I don't know how I could be this confused, indeed only the boundary is necessary.
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