complex numbers - 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

$$\arcsin(x)=\frac1i\log\left(iz+\sqrt{1-z^2}\right)$$