Hyperbolic Cotangent of Complex Number/Formulation 3
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Theorem
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
- $\map \coth {a + b i} = \dfrac {\coth a + \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\cot b - \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i$
where:
- $\cot$ denotes the real cotangent function
- $\coth$ denotes the hyperbolic cotangent function.
Proof
\(\ds \map \coth {a + b i}\) | \(=\) | \(\ds \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}\) | Hyperbolic Cotangent of Complex Number: Formulation 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - i \coth a \cot b} \paren {\coth a + i \cot b} } {\paren {\coth a - i \cot b} \paren {\coth a + i \cot b} }\) | multiplying denominator and numerator by $\coth a + i \cot b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\paren {1 - i \coth a \cot b} \paren {\coth a + i \cot b} } {\coth^2 a + \cot^2 b}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\coth a + i \cot b - i \coth^2 a \cot b + \coth a \cot^2 b} {\coth^2 a + \cot^2 b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\coth a + \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\cot b - \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i\) |
$\blacksquare$