Inverse Tangent of Imaginary Number

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Theorem

$\map {\tan^{-1} } {i x} = i \tanh^{-1} x$


Proof 1

\(\ds y\) \(=\) \(\ds \map {\tan^{-1} } {i x}\)
\(\ds \leadsto \ \ \) \(\ds \tan y\) \(=\) \(\ds i x\) Definition of Inverse Tangent
\(\ds \leadsto \ \ \) \(\ds i \tan y\) \(=\) \(\ds - x\) $i^2 = -1$
\(\ds \leadsto \ \ \) \(\ds \map \tanh {i y}\) \(=\) \(\ds -x\) Tangent in terms of Hyperbolic Tangent
\(\ds \leadsto \ \ \) \(\ds i y\) \(=\) \(\ds \map {\tanh^{-1} } {-x}\) Definition of Inverse Hyperbolic Tangent
\(\ds \leadsto \ \ \) \(\ds i y\) \(=\) \(\ds -\tanh^{-1} x\) Inverse Hyperbolic Tangent is Odd Function
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds i \tanh^{-1} x\) multiplying both sides by $-i$

$\blacksquare$


Proof 2

\(\ds \map {\tan^{-1} } {i x}\) \(=\) \(\ds \frac i 2 \map \ln {\frac {1 - i \paren {i x} } {1 + i \paren {i x} } }\) Arctangent Logarithmic Formulation
\(\ds \) \(=\) \(\ds \frac i 2 \map \ln {\frac {1 + x} {1 - x} }\)
\(\ds \) \(=\) \(\ds i \tanh^{-1} x\) Definition of Inverse Hyperbolic Tangent

$\blacksquare$


Sources