Sine of Complex Number

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Theorem

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

Let $i$ be the imaginary unit.

Then:

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

where:

$\sin$ denotes the sine function (real and complex)
$\cos$ denotes the real cosine function
$\sinh$ denotes the hyperbolic sine function
$\cosh$ denotes the hyperbolic cosine function.


Proof 1

\(\ds \sin \paren {a + b i}\) \(=\) \(\ds \sin a \cos \paren {b i} + \cos a \sin \paren {b i}\) Sine of Sum
\(\ds \) \(=\) \(\ds \sin a \cosh b + \cos a \sin \paren {b i}\) Hyperbolic Cosine in terms of Cosine
\(\ds \) \(=\) \(\ds \sin a \cosh b + i \cos a \sinh b\) Hyperbolic Sine in terms of Sine

$\blacksquare$


Proof 2

\(\ds \sin a \cosh b + i \cos a \sinh b\) \(=\) \(\ds \frac {e^{i a} - e^{-i a} } {2 i} \frac {e^b - e^{-b} } 2 + i \frac {e^{i a} + e^{-i a} } 2 \frac {e^b - e^{-b} } 2\) Euler's Sine Identity, Euler's Cosine Identity, Definition of Hyperbolic Sine, Definition of Hyperbolic Cosine
\(\ds \) \(=\) \(\ds \frac {e^{b + i a} - e^{-b + i a} - e^{b - i a} + e^{-b - i a} - e^{b + i a} + e^{-b + i a} - e^{b - i a} + e^{-b - i a} } {4 i}\) simplifying
\(\ds \) \(=\) \(\ds \frac {e^{-b - i a} - e^{b - i a} } {2 i}\) simplifying
\(\ds \) \(=\) \(\ds \frac {e^{i \paren {a + b i} } - e^{-i \paren {a + b i} } } {2 i}\)
\(\ds \) \(=\) \(\ds \map \sin {a + b i}\) Euler's Sine Identity

$\blacksquare$


Also see