# Solving complex number equations involving trig

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?

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 $$u$$.

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

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