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Math Help - interesting complex equation

  1. #1
    Junior Member
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    interesting complex equation

    Here is an interesting problem I'm having some difficulty with:
    Find all the complexes z that satisfy the equation:
    cos(z) = cosh(z).
    The answer should be z=k(pi)(1+i) & z=k(pi)(1-i), but I don't know how to show this.
    Any help, folks?
    Thanks.
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  2. #2
    Super Member
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    Salam

    you must use the two following identities.

    \cosh z = \cos i z

    \cos A = \cos B \ \ \ \rightarrow A = 2 \pi n \pm B

    Can you take it form here, if you need help proving those identities just ask.

    Bobak
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  3. #3
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    hi bobak,
    thanks a lot for the help. only i thought cosA = cosB => A = 2n(pi) +/- B holds true for A & B real numbers. how does this also work for complexes. is cos a periodic function for complexes.
    can you please clarify?
    thanks.
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  4. #4
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    Quote Originally Posted by tombrownington View Post
    hi bobak,
    thanks a lot for the help. only i thought cosA = cosB => A = 2n(pi) +/- B holds true for A & B real numbers. how does this also work for complexes. is cos a periodic function for complexes.
    can you please clarify?
    thanks.
    consider the function \cos z = \frac{e^{iz} + e^{-iz} }{2} adding 2 \pi to z will make no difference as \cos (z +2 \pi) = \frac{e^{i(z +2 \pi)} + e^{-i(z +2 \pi)} }{2} \ = \  \frac{e^{iz}e^{i 2\pi} + e^{-iz}e^{-i 2\pi} }{2} however e^{i 2\pi} = 1 so \cos z = \cos (z + 2 \pi).

    Bobak
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