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