Hyperbolic Tangent of Complex Number/Formulation 3
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Theorem
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
- $\tanh \paren {a + b i} = \dfrac {\tanh a + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b} + \dfrac {\tan b - \tanh^2 a \tan b} {1 + \tanh^2 a \tan^2 b} i$
where:
- $\tan$ denotes the real tangent function
- $\tanh$ denotes the hyperbolic tangent function.
Proof
| \(\ds \tan \paren {a + b i}\) | \(=\) | \(\ds \dfrac {\tanh a + i \tan b} {1 + i \tanh a \tan b}\) | Hyperbolic Tangent of Complex Number: Formulation 2 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\tanh a + i \tan b} \paren {1 - i \tanh a \tan b} } {1 + \tanh^2 a \tan^2 b}\) | multiplying denominator and numerator by $1 - i \tanh a \tan b$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tanh a + i \tan b - i \tanh^2 a \tan b + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tanh a + \tanh a \tan^2 b} {1 + \tanh^2 a \tan^2 b} + \frac {\tan b - \tanh^2 a \tan b} {1 + \tanh^2 a \tan^2 b} i\) |
$\blacksquare$