prove the following equation about inverse of tan in logarithmic for $$\arctan(z)=\frac1{2i}\log\left(\frac{1+iz}{1-iz}\right)$$
i have tried but my answer doesn't matches to the equation .the componendo dividendo property might have been used. where
$$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$
 A: Using Euler's formula:-
$$e^{i\theta}=\cos\theta+i\sin\theta$$
we have
$$2i\sin\theta=e^{i\theta}-e^{-i\theta}\text{ (1)}$$
$$2\cos\theta=e^{i\theta}+e^{-i\theta}\text{ (2)}$$
Dividing Equation $(1)$ by $(2)$ and then dividing both sides by $i$ results in
$$\tan\theta=\frac{1}{i}\left(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\right)$$ 
Let $z=\tan\theta \Leftrightarrow \theta=\arctan z$, so that
$$z=\frac{1}{i}\left(\frac{e^{i\theta}-e^{-i\theta}}{e^{i\theta}+e^{-i\theta}}\right)\\\color{blue}{\Rightarrow iz(e^{i\theta}+e^{-i\theta})=e^{i\theta}-e^{-i\theta}}\\\Rightarrow e^{i\theta}(1-iz)=e^{-i\theta}(1+iz)
\\\Rightarrow e^{2i\theta}=\frac{1+iz}{1-iz}\\\Rightarrow 2i\theta=\log\left(\frac{1+iz}{1-iz}\right)\\\Rightarrow \theta=\frac{1}{2i}\log\left(\frac{1+iz}{1-iz}\right)\\\Rightarrow \arctan z=\frac{1}{2i}\log\left(\frac{1+iz}{1-iz}\right)$$ 
A: Let $\arctan(z) = t$. We then have $z = \tan(t)$. Now look at the right hand side and we have,
$$\dfrac1{2i} \log\left(\dfrac{1+iz}{1-iz}\right) = \dfrac1{2i} \log\left(\dfrac{1+i\tan(t)}{1-i\tan(t)}\right) = \dfrac1{2i} \log\left(\dfrac{\cos(t)+i\sin(t)}{\cos(t)-i\sin(t)}\right) = \dfrac1{2i} \log\left( \dfrac{e^{it}}{e^{-it}}\right) = t$$
