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