Inverse Hyperbolic Tangent/Examples/i

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Example of Inverse Hyperbolic Tangent

$\tanh^{-1} \paren i = \dfrac {\paren {4 k + 1} \pi i} 4$


Proof

By definition of inverse hyperbolic tangent:

$\tanh^{-1} \paren i := \set {z \in \C: \tanh z = i}$


Thus:

\(\ds \tanh z\) \(=\) \(\ds i\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {\exp \paren {2 z} - 1} {\exp \paren {2 z} + 1}\) \(=\) \(\ds i\) Definition 3 of Hyperbolic Tangent
\(\ds \leadsto \ \ \) \(\ds \exp \paren {2 z} - 1\) \(=\) \(\ds i \paren {\exp \paren {2 z} + 1}\) rearranging
\(\ds \) \(=\) \(\ds i \exp \paren {2 z} + i\) rearranging
\(\ds \leadsto \ \ \) \(\ds \exp \paren {2 z} \paren {1 - i}\) \(=\) \(\ds 1 + i\)
\(\ds \leadsto \ \ \) \(\ds \exp \paren {2 z}\) \(=\) \(\ds \dfrac {1 + i} {1 - i}\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds \dfrac 1 2 \ln \dfrac {1 + i} {1 - i}\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \ln \dfrac {\paren {1 + i}^2 } {\paren {1 - i} \paren {1 + i} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \ln \dfrac {\paren {2i} } 2\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \ln i\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \paren {\paren {4 k + 1} \dfrac {\pi i} 2}\) Natural Logarithm of $i$
\(\ds \) \(=\) \(\ds \dfrac {\paren {4 k + 1} \pi i} 4\) for all $k \in \Z$

$\blacksquare$


Sources

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