What is $\tan ^{-1} (5i/3)$ 
What is $\tan ^{-1} (5i/3)$

My progress: Let $\tan x= \dfrac{5i}{3}= \dfrac{\sin x}{\cos x}$
I tried using $\sin x= \dfrac{e^{ix}-e^{-ix}}{2}, \cos x= \dfrac{e^{ix}+e^{-ix}}{2}$ to show that $\dfrac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}= \dfrac{-5}{3}$ or $e^{2ix}= \dfrac{-1}{4}$, but I'm stuck here.

Let us write $z =\frac{5i}{3}$ and also $w =\arctan (\frac{5i}{3}) =\arctan z$. Hence $\tan w=z$ and we want to solve for $w$. We have $\sin w= \frac{e^{iw} -e^{-iw}}{2i}$ and $\cos w=\frac{e^{iw}+e^{-iw}}{2}$.  So then we have $\tan w=\frac{e^{iw}- e^{-iw}}{i(e^{iw}+e^{-iw})}$.  
Let $p =e^{iw}$ so that $\frac{1}{p} = e^{-iw}$ then we have $$\frac{p-\frac{1}{p}}{i(p+\frac{1}{p})} = \tan w=z$$ where for now we have $z= \frac{5i}{3}$. We can simplify this to $$\frac{p^2-1}{p^2+1} = z$$ $$\Rightarrow p^2-1 = iz(p^2 +1)$$ $$\Rightarrow (1-iz)p^2 = (1+ iz)$$ $$\Rightarrow p^2 = \frac{1+iz}{1-iz}$$ $$\Rightarrow e^{2iw} = \frac{1+iz}{1-iz}$$ $$ w = \frac{1}{2i} \times \ln(\frac{1+iz}{1-iz})$$ Now substitute for $z=\frac{5i}{3}$ for the answer.
