Tangent of Complex Number/Formulation 1

Theorem

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

Let $i$ be the imaginary unit.

Then:

$\tan \paren {a + b i} = \dfrac {\sin a \cosh b + i \cos a \sinh b} {\cos a \cosh b - i \sin a \sinh b}$

where:

$\tan$ denotes the complex tangent function

$\sin$ denotes the real sine function

$\cos$ denotes the real cosine function

$\sinh$ denotes the hyperbolic sine function

$\cosh$ denotes the hyperbolic cosine function.

Proof

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

\(=\)

\(\ds \frac {\sin \paren {a + b i} } {\cos \paren {a + b i} }\)

Definition of Complex Tangent Function

\(\ds \)

\(=\)

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

Sine of Complex Number and Cosine of Complex Number

$\blacksquare$

Also see

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