Hyperbolic Cotangent of Complex Number/Formulation 2

Theorem

Let $a$ and $b$ be real numbers.

Let $i$ be the imaginary unit.

Then:

$\map \coth {a + b i} = \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}$

where:

$\ds \map \coth {a + b i}$

$=$

$\ds \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}$

$\ds $

$=$

$\ds \dfrac {\coth a \cos b + i \sin b} {\cos b + i \coth a \sin b}$

multiplying denominator and numerator by $\dfrac 1 {\sinh a}$

$\ds $

$=$

$\ds \dfrac {\coth a \cot b + i} {\cot b + i \coth a}$

multiplying denominator and numerator by $\dfrac 1 {\sin b}$

$\ds $

$=$

$\ds \dfrac {1 - i \coth a \cot b} {\coth a - i \cot b}$

multiplying denominator and numerator by $-i$

$\blacksquare$