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Showing a complicated mess is equal to cot(z)

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the function f(z) whose real part is

    [tex]\frac{sin(x)}{cosh(2y)-cos(2x)}[/tex]

    where z = x + iy

    2. Relevant equations

    I already know f(z)=cot(z), using FullSimplify[] in mathematica on the expression I get for u + iv when I apply the Cauchy-Reimann equations.

    More explicitly,

    f(x,y) = u(x,y)+iv(x,y) = [tex]\frac{-sin(2x)+isinh(2y)}{cos(2x)-cosh(2y)}[/tex]


    3. The attempt at a solution

    I've tried to derive without mathematica that this is cot(z) about a thousand different ways. I can't seem to manipulate the expression into a usable form. Am I missing something obvious?

    Help!!!

    Thanks!!

    -Nathan
     
  2. jcsd
  3. Sep 1, 2008 #2

    HallsofIvy

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    There can exist an infinite number of functions having that real part. Do you mean a function analytic on the entire complex plane?
     
  4. Sep 1, 2008 #3
    Yes, a function analytic on a certain subset of the complex plane.
     
  5. Sep 1, 2008 #4

    tiny-tim

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    Hi Nathan! :smile:

    Hint:

    [tex]\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}[/tex] = … ? :smile:
     
  6. Sep 1, 2008 #5
    I'm sorry, Tiny Tim, I don't see how that helps me:

    I get,

    [tex]
    \frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)} = -(Cos(2z)+Sin(2z))
    [/tex]

    What does that have to do with reducing what I found above?
     
  7. Sep 1, 2008 #6

    tiny-tim

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    Hi Nathan! :smile:

    How did you get that? :confused:

    I get [tex]\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)}[/tex] = … ? :smile:
     
  8. Sep 1, 2008 #7
    Thanks for the help, tiny-tim

    I think I dropped an i

    [tex]
    \frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)} = sin^2(z) - cos^2(z) - 2isin(z)cos(z) = -(cos(2z) + i sin(2z)) = -e^{2iz}
    [/tex]

    But I still don't see how this helps me with my original problem...
     
  9. Sep 1, 2008 #8

    tiny-tim

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    hmm … quicker would be …
    [tex]\frac{1\ -\ i\,cot(z)}{1\ +\ i\,cot(z)}\ =\ \frac{sin(z)\ -\ i\,cos(z)}{sin(z)\ +\ i\,cos(z)}\ =\ -\frac{i\,e^{iz}}{i\,e{-iz}}\ =\ -e^{2iz}[/tex]
    You wanted to prove that cot(z) = [tex]\frac{-sin(2x)+isinh(2y)}{cos(2x)-cosh(2y)}[/tex] …

    just plug that into the above formula. :wink:
     
    Last edited: Sep 1, 2008
  10. Sep 1, 2008 #9
    Wow, that was a really hard problem...

    Thanks, for the help, tiny-tim!

    If you wouldn't mind my asking, how did you come up with that?
     
  11. Sep 1, 2008 #10

    tiny-tim

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    erm … I'm familiar with (1 + tanhz)/(1 - tanhz) = e2z

    (as you should be! :wink:)

    so I just used iz instead of z. :smile:
     
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