Arctangent Logarithmic Formulation

Theorem
For any real number $x$,


 * $ \displaystyle \arctan x = \frac 1 2 i \ln \left({ \frac{1-ix}{1+ix} }\right) $

where $\arctan x$ is the arctangent and $i^2 = -1$.

Proof
Assume $ y \in \R $, $ -\frac \pi 2 \le y \le \frac \pi 2 $.