Secant of Complex Number

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Theorem

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

Let $i$ be the imaginary unit.


Then:

$\sec \paren {a + b i} = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}$

where:

$\sec$ denotes the complex secant 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

\(\ds \sec \paren {a + b i}\) \(=\) \(\ds \frac 1 {\cos \paren {a + b i} }\) Definition of Complex Secant Function
\(\ds \) \(=\) \(\ds \dfrac 1 {\cos a \cosh b - i \sin a \sinh b}\) Cosine of Complex Number
\(\ds \) \(=\) \(\ds \dfrac {\cos a \cosh b + i \sin a \sinh b} {\paren {\cos a \cosh b + i \sin a \sinh b} \paren {\cos a \cosh b - i \sin a \sinh b} }\) multiplying denominator and numerator by $\cos a \cosh b + i \sin a \sinh b$
\(\ds \) \(=\) \(\ds \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b - i^2 \sin^2 a \sinh^2 b}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}\) Definition of Imaginary Unit

$\blacksquare$


Also see