Hyperbolic Secant of Complex Number
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Theorem
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
- $\sech \paren {a + b i} = \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}$
where:
- $\sech$ denotes the hyperbolic 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
\(\ds \sech \paren {a + b i}\) | \(=\) | \(\ds \frac 1 {\cosh \paren {a + b i} }\) | Definition of Hyperbolic Secant | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {\cosh a \cos b + i \sinh a \sin b}\) | Hyperbolic Cosine of Complex Number | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\cosh a \cos b - i \sinh a \sin b} {\paren {\cosh a \cos b + i \sinh a \sin b} \paren {\cosh a \cos b - i \sinh a \sin b} }\) | multiplying denominator and numerator by $\cosh a \cos b - i \sinh a \sin b$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b - i^2 \sinh^2 a \sin^2 b}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\cosh a \cos b - i \sinh a \sin b} {\cosh^2 a \cos^2 b + \sinh^2 a \sin^2 b}\) | Definition of Imaginary Unit |
$\blacksquare$