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:
- $\tan \paren {a + b i} = \dfrac {\tan a - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b} + \dfrac {\tanh b + \tan ^2 a \tanh b} {1 + \tan ^2 a \tanh ^2 b} i$
where:
- $\tan$ denotes the tangent function (real and complex)
- $\tanh$ denotes the hyperbolic tangent function.
Proof
| \(\ds \tan \paren {a + b i}\) | \(=\) | \(\ds \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}\) | Tangent of Complex Number: Formulation 2 | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\tan a + i \tanh b} \paren {1 + i \tan a \tanh b} } {1 + \tan ^2 a \tanh ^2 b}\) | multiplying denominator and numerator by $1 + i \tan a \tanh b$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tan a + i \tanh b + i \tan ^2 a \tanh b - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\tan a - \tan a \tanh ^2 b} {1 + \tan ^2 a \tanh ^2 b} + \frac {\tanh b + \tan ^2 a \tanh b} {1 + \tan ^2 a \tanh ^2 b} i\) |
$\blacksquare$