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Math Help - Urgent advice please on a complex no question:

  1. #1
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    Urgent advice please on a complex no question:

    Let B = {z:|z|<3, Re z + Im z <= 0}

    prove that the function f(z) = cosh(1+iz^2)/cos z

    is bounded on the set delta B

    Could anyone give me a steer on how I should go about solving this?


    Many thanks
    Graham
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  2. #2
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    Quote Originally Posted by grahamlee View Post
    Let B = {z:|z|<3, Re z + Im z <= 0}

    prove that the function f(z) = cosh(1+iz^2)/cos z

    is bounded on the set delta B

    Could anyone give me a steer on how I should go about solving this?


    Many thanks
    Graham
    Your question seems weird to me; a typo, maybe?

    The value -\frac{\pi}{2} is in B, right? It is such that \cos\left(-\frac{\pi}{2}\right)=0, while \cosh(1+i\left(-\frac{\pi}{2}\right)^2) has no reason to be equal to 0, so that f diverges at -\frac{\pi}{2}, hence f is not even defined on B...
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  3. #3
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    I'm not sure to be honest. The approach I have taken is as follows:

    |cosh(1+iz^2)/cos z|

    = |1/2 (e^(1+iz^2)+e^-(1+iz^2)) / 1/2(e^iz+e^-iz)|

    = e^1+|z^2|+e^1+|z^2| / |e^z|+|e^z|

    = e^1+|z^2| / e^|z|

    When |z|< 3, this equates to e^7

    It looks right but I can't find a way to check this.
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  4. #4
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    Quote Originally Posted by grahamlee View Post
    I'm not sure to be honest. The approach I have taken is as follows:

    |cosh(1+iz^2)/cos z|

    = |1/2 (e^(1+iz^2)+e^-(1+iz^2)) / 1/2(e^iz+e^-iz)|

    = e^1+|z^2|+e^1+|z^2| / |e^z|+|e^z|

    = e^1+|z^2| / e^|z|

    When |z|< 3, this equates to e^7

    It looks right but I can't find a way to check this.
    It isn't right. Most notably (I assume you mean \leq instead of =): you can't use the inequality |e^{iz}+e^{-iz}|\leq |e^{iz}|+|e^{-iz}| at the denominator since it would give you an upper bound. And again, look at what happens when z=-\frac{\pi}{2}: the denominator is 0, so you can't expect to bound it away from 0.
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