Complex numbers hyperbolic trig

  • #1
187
0
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
 

Answers and Replies

  • #2
121
1
The question is correct. Remember i^2 = -1.
 
  • #3
187
0
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:
  • #4
187
0
the second part of the second equation there should be e^(iu)+e^(-iu)/2
 
  • #5
121
1
[tex] tan u = i(e^{-iu} - e^{iu}) / (e^{-iu} + e^{iu}) [/tex]
 

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