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