Tangent of Complex Number/Formulation 2

Theorem

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

Let $i$ be the imaginary unit.

Then:

$\tan \paren {a + b i} = \dfrac {\tan a + i \tanh b} {1 - i \tan a \tanh b}$

where:

$\tan$ denotes the tangent function (real and complex)

$\tanh$ denotes the hyperbolic tangent function.

Proof

\(\ds \tan \paren {a + b i}\)

\(=\)

\(\ds \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}\)

Tangent of Complex Number: Formulation 1

\(\ds \)

\(=\)

\(\ds \dfrac {\tan a \cosh b + i \sinh b} {\cosh b - i \tan a \sinh b}\)

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

\(\ds \)

\(=\)

\(\ds \frac {\tan a + i \tanh b} {1 - i \tan a \tanh b}\)

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

$\blacksquare$

Also see

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