Inverse Tangent of Imaginary Number

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Theorem

$\tan^{-1} \left({i x}\right) = i \tanh^{-1} x$


Proof 1

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

$\blacksquare$


Proof 2

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

$\blacksquare$


Sources