Secant of Complex Number
Jump to navigation
Jump to search
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 |
$\blacksquare$