Complex numbers hyperbolic trig

[thenewbosco]

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

 

[Max Eilerson]

The question is correct. Remember i^2 = -1.

 

[thenewbosco]

what i have done is:
for the left side:

$$tanh u = \frac{sinh u}{cosh u} = \frac{e^{iu}-e^{-iu}}{e^{iu}+e^{-iu}}$$

but then right side

$$i tan u = i\frac{ \frac{e^u-e^{-u}}{2i} } \frac{ e^{iu}+e^{-iu} } {2} }=<br /> \frac{e^{u}-e^{-u}}{e^{iu}+e^{-iu}}$$

what have i done wrong here?

 

[thenewbosco]

the second part of the second equation there should be e^(iu)+e^(-iu)/2

 

[Max Eilerson]

$$tan u = i(e^{-iu} - e^{iu}) / (e^{-iu} + e^{iu})$$