Arctangent of Imaginary Number

Theorem
Let $x$ belong to the open real interval $\openint {-1} 1$.

Then:
 * $\map {\tan^{-1} } {i x} = \dfrac i 2 \map \ln {\dfrac {1 + x} {1 - x} }$

where $\tan$ is the complex tangent function, $\ln$ is the real natural logarithm, and $i$ is the imaginary unit.

Proof
Let $y = \map {\tan^{-1} } {i x}$.

Let $x = \tanh \theta$, then $\theta = \tanh^{-1} x$.