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Inverse tangent of a complex number

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to find ##\tan^{-1}(2i)##.


    2. Relevant equations



    3. The attempt at a solution
    So far I have ##\tan^{-1}(2i)=z\iff tan z= 2i\iff \dfrac{sin z}{cos z}=2i ##. From here I get that
    ##-3=e^{-2zi}##. I do no know how to take it further to get ##z=i\dfrac{\ln 3}{2}+(\dfrac{\pi}{2}+\pi n)##. Should I use natural logarithms but the problem is that I have a ##-3## which wont allow me to take the natural log of both sides. Any help would be great thanks.
     
  2. jcsd
  3. Mar 29, 2014 #2

    haruspex

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    I get -(3+4i)/5 for that.
    The log of a negative number is fine when complex answers are allowed. It's only 0 that has no log in the complex plane.
     
  4. Mar 29, 2014 #3

    Curious3141

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    The solution is as simple as using ##\tan{ix} = itanh{x}##.

    But you can use your method and do the algebra, it's just a little more work. Comes to the same answer.
     
    Last edited: Mar 29, 2014
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