Hyperbolic Cotangent of Complex Number/Formulation 3

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

\(\ds \)

\(=\)

\(\ds \dfrac {\coth a + i \cot b - i \coth^2 a \cot b + \coth a \cot^2 b} {\coth^2 a + \cot^2 b}\)

\(\ds \)

\(=\)

\(\ds \dfrac {\coth a + \coth a \cot^2 b} {\coth^2 a + \cot^2 b} + \dfrac {\cot b - \coth^2 a \cot b} {\coth^2 a + \cot^2 b} i\)

$\blacksquare$

Also see

• Hyperbolic Sine of Complex Number
• Hyperbolic Cosine of Complex Number
• Hyperbolic Tangent of Complex Number
• Hyperbolic Cosecant of Complex Number
• Hyperbolic Secant of Complex Number