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Exp(tanz) = 1, complex analysis

  1. Jun 15, 2007 #1

    malawi_glenn

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

    Find all solutions to:

    [tex]e^{\tan z} =1, z\in \mathbb{C}[/tex]


    2. Relevant equations

    z = x+yi

    [tex]\log z = ln|z| + iargz +2\pihi, h\in \mathbb{Z}[/tex]


    [tex]\log e^{z} = x + iy +2\pihi, h\in \mathbb{Z}[/tex]


    [tex]Log e^{z} = x + iy[/tex]

    3. The attempt at a solution

    I do not really know how to approach this, I tried to beging with writing tan(z) as Alots of cos(x)sinh(y) etc..

    But can I do the Log(e^tan(z)) at the left side, then do the log(1) at the right side? I mean the "only" difference is that you get this [tex]2\pihi, h\in \mathbb{Z}[/tex] on both sides, so you always reduce both these terms to one: [tex]2\piUi, U\in \mathbb{Z}[/tex]

    What do you think?
     
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  3. Jun 15, 2007 #2

    Dick

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    Ok, taking logs you get tan(z)=2*pi*i*n (yes, no need for separate 2*pi*i*n's on both sides. Why not write tan(z) in terms of complex exponentials, let t=e^(iz) and solve the resulting quadratic for t? Guess I'm not sure where you are having problems.
     
  4. Jun 15, 2007 #3

    malawi_glenn

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    so if I write

    [tex] \tan z =\dfrac{1}{i} \dfrac{t-t^{-1}}{t+t^{-1}} = \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1} [/tex]

    then perform the Log(e^tanz) on left side, then log(1) on right side, is that a "legal" act? or do I have to take log - log /// Log - Log ?..
     
    Last edited: Jun 15, 2007
  5. Jun 15, 2007 #4

    malawi_glenn

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    or you mean AFTER taking log's i write tanz as that?
     
  6. Jun 15, 2007 #5

    malawi_glenn

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    okay this is as far I can get:

    [tex] e^{iz} = t [/tex]

    [tex] z = - i \ln |t| + argt[/tex]

    [tex]
    \tan z = \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1}; t^{2} \neq -1; \Rightarrow t \neq \pm i
    [/tex]

    [tex] \tan z = \log 1 = 0 +0i + 2\pi ni; n \in \mathbb{Z}[/tex]


    [tex]\Rightarrow \dfrac{1}{i} \dfrac{t^{2}-1}{t^{2}+1} = 2\pi ni[/tex]

    [tex]\Rightarrow \dfrac{t^{2}-1}{t^{2}+1} = -2\pi n[/tex]

    [tex]\Rightarrow t^{2}-1 = -2\pi n(t^{2}+1) [/tex]

    ...

    [tex]\Rightarrow t = \pm \sqrt{\dfrac{2\pi n-1}{2\pi n+1}} \neq -1[/tex]

    [tex]\Rightarrow n \neq 0 [/tex]

    [tex] t_{1} = "+"sqrt(.. \Rightarrow arg(t_{1}) = 0 + 2\pi k, k \in \mathbb{Z}[/tex]
    [tex] t_{2} = "-"sqrt(.. \Rightarrow arg(t_{2}) = \pi + 2\pi k[/tex]

    it can be shown that t is always real, if n = -2, we get a negative nominator and a negative denominator, hence the number inside the squareroot is always positive.

    [tex] z_{1} = - \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + \pi + 2\pi k[/tex]

    [tex] z_{2} = - \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + 2\pi k[/tex]

    Answer in textbook:

    [tex] z = k\pi [/tex]

    and:

    [tex] z = \dfrac{i}{2} \ln \left( \dfrac{2\pi n-1}{2\pi n+1} \right) + \frac{\pi}{2} + 2\pi k[/tex]

    LOL I am soooo cloose!! =(
     
    Last edited: Jun 15, 2007
  7. Jun 15, 2007 #6

    Dick

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    I get t^2=(1-2*pi*n)/(1+2*pi*n). So I get a number that is always negative. So I agree with the pi/2 phase in the book answer. But I get +/-pi/2+2*pi*k which is the same thing as pi/2+pi*k. But I'm also getting -i/2 in front of the log. (But I seem to get the right answer with either sign - this is troubling me).
     
  8. Jun 15, 2007 #7

    malawi_glenn

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    yes I did a relly sucky thing in getting the t^2.. I should be ashamed! :)


    if n = 0 then it is positive, right? But can not be zero according to the aswer, now why is that?
     
    Last edited: Jun 15, 2007
  9. Jun 15, 2007 #8

    Dick

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    Aggghhh. And get this, the sign in front of the log doesn't matter either. Take for example the cases n=1 and n=-1. The arguments of the logs are just reciprocals of each other. Must be fun to create obfuscated answers for these things.
     
  10. Jun 15, 2007 #9

    Dick

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    n=0 gives you the real solutions, the k*pi list. Not to hard so see working it as a separate case.
     
  11. Jun 15, 2007 #10

    malawi_glenn

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    I got it now, thanx again for all the help. I hope i did not made you indignant.
     
  12. Jun 15, 2007 #11

    Dick

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    I was working to hard to correct my own blunders to become indignant over yours. :smile:
     
  13. Jun 15, 2007 #12

    malawi_glenn

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    I dont understand why the answers must be so "clean".. it only cunfuses and does not get you learn more about the actual subject.
     
  14. Jun 16, 2007 #13

    Dick

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    It can be helpful to numerically evaluate a few cases of a solution for small n,k etc and then compare with yours. It's pretty clear when they don't agree - and if they do it can help to make it clearer why.
     
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