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Complex numbers hyperbolic trig

  1. Oct 28, 2006 #1
    it says to use exponentials to prove:

    tanh (iu) = i tan u

    however i do not get the correct relationship, is this an error in the question perhaps
     
  2. jcsd
  3. Oct 28, 2006 #2
    The question is correct. Remember i^2 = -1.
     
  4. Oct 28, 2006 #3
    what i have done is:
    for the left side:

    [tex]tanh u = \frac{sinh u}{cosh u} = \frac{e^{iu}-e^{-iu}}{e^{iu}+e^{-iu}}[/tex]

    but then right side

    [tex] i tan u = i\frac{ \frac{e^u-e^{-u}}{2i} } \frac{ e^{iu}+e^{-iu} } {2} }=
    \frac{e^{u}-e^{-u}}{e^{iu}+e^{-iu}}[/tex]

    what have i done wrong here?
     
    Last edited: Oct 28, 2006
  5. Oct 28, 2006 #4
    the second part of the second equation there should be e^(iu)+e^(-iu)/2
     
  6. Oct 28, 2006 #5
    [tex] tan u = i(e^{-iu} - e^{iu}) / (e^{-iu} + e^{iu}) [/tex]
     
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