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 $-i$

$\blacksquare$


Also see