Arctangent of Imaginary Number
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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 |
$\blacksquare$