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Inverse tangent function in real and complex domain

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    See attached file.


    2. Relevant equations



    3. The attempt at a solution
    I've only been able to do part (a) of this question.
    I ended up with:
    [tex]
    tanz= i ({\frac{1-e^{(2iz)}}{1+e^{(2iz)}}})
    [/tex]
    I'm not sure how to approach the next two parts. If anyone could give me any pointers, I'd be very grateful! Thanks!
     

    Attached Files:

  2. jcsd
  3. May 12, 2012 #2
    Why not just for starters approach it formally, get the answer, then justify what you did. I'll use the alternate formula:

    [tex]\arctan(z)=i/2\log \frac{i+z}{i-z}[/tex]
    and you have:
    [tex]tan(w)=i\left(\frac{1-e^{2iw}}{1+e^{2iw}}\right)[/tex]
    ok then, how about if I let:
    [tex]w=i/2\log\frac{i+z}{i-z}[/tex]
    then can you not just muscle-through:
    [tex]\tan(w)=\tan\left(i/2\log\frac{i+z}{i-z}\right)[/tex]
     
    Last edited: May 12, 2012
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