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

\(\ds \tan y\)

\(=\)

\(\ds i x\)

\(\ds \tan y\)

\(=\)

\(\ds i \tanh \theta\)

\(\ds \tan y\)

\(=\)

\(\ds \map \tan {i \theta}\)

Hyperbolic Tangent in terms of Tangent

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds i \theta\)

\(\ds y\)

\(=\)

\(\ds i \tanh^{-1} x\)

\(\ds y\)

\(=\)

\(\ds \frac i 2 \map \ln {\frac {1 + x} {1 - x} }\)

Definition of Real Hyperbolic Arctangent

$\blacksquare$