Cotangent of Complex Number/Formulation 3

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$

Also see

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