Solving complex number equations involving trig

Question

Let $w$ be a complex number. By solving the equation $\frac{u^2 - 1}{u^2 + 1} = iw$ for a suitable complex number $u$, find an expression for $\tan^{-1}(w)$.

I did the first part and got that $u^2 = \frac{1 + iw}{1 - iw}$ but have no idea where to go from here. Should I be integrating or something?

Answer 1

This is kind of a neat one.

$$ \frac{u^2 - 1}{u^2 + 1} = iw $$

$$ \frac{ \frac{u - 1/u }{2i} }{ \frac{u + 1/u }{2} }= w $$

The definition of sine and cosine can be used with the right choice of $u$.

$$ u = e^{i\theta} $$

$$ \frac{ \frac{e^{i\theta} - e^{-i\theta} }{2i} }{ \frac{e^{i\theta} + e^{-i\theta} }{2} } = w $$

$$ \frac{ \sin(\theta)} {\cos(\theta)} = \tan(\theta)= w $$

Finally, the expression can be converted back into terms of $u$.

$$ \tan^{-1}(w)= \theta = \frac{\ln(u)}{i} = -\ln(u)i $$

Answer 2

Let $\gamma=\tan ^{-1} w$ then we have $$w=\tan \gamma =(1/i)(\frac {e^{2i \gamma }-1}{e^{2i \gamma}+1}) $$

Thus with $u=e^{i\gamma} $we have $$ \tan ^{-1} w=\frac {\ln u}{i}$$