Arctangent of Imaginary Number

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Theorem

Let $x$ be a real number.

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

\(\displaystyle \tan y\) \(=\) \(\displaystyle i x\)
\(\displaystyle \tan y\) \(=\) \(\displaystyle i \tanh \theta\)
\(\displaystyle \tan y\) \(=\) \(\displaystyle \map \tan {i \theta}\) Hyperbolic Tangent in terms of Tangent
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle i \theta\)
\(\displaystyle y\) \(=\) \(\displaystyle i \tanh^{-1} x\)
\(\displaystyle y\) \(=\) \(\displaystyle \frac i 2 \map \ln {\frac {1 + x} {1 - x} }\) Definition of Real Hyperbolic Arctangent

$\blacksquare$