# Sine Exponential Formulation

## Theorem

For any complex number $z$:

$\sin z = \dfrac {\map \exp {i z} - \map \exp {-i z} } {2 i}$
$\exp z$ denotes the exponential function
$\sin z$ denotes the complex sine function
$i$ denotes the inaginary unit.

### Real Domain

This result is often presented and proved separately for arguments in the real domain:

$\sin x = \dfrac {e^{i x} - e^{-i x} } {2 i}$

## Proof 1

Recall the definition of the sine function:

 $\ds \sin z$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$ $\ds$ $=$ $\ds z - \frac {z^3} {3!} + \frac {z^5} {5!} - \frac {z^7} {7!} + \cdots + \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!} + \cdots$

Recall the definition of the exponential as a power series:

 $\ds \exp z$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$ $\ds$ $=$ $\ds 1 + \frac z {1!} + \frac {z^2} {2!} + \frac {z^3} {3!} + \cdots + \frac {z^n} {n!} + \cdots$

Then, starting from the right hand side:

 $\ds \frac {\exp \paren {i z} - \exp \paren {-i x} } {2 i}$ $=$ $\ds \frac 1 {2 i} \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} - \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }$ $\ds$ $=$ $\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n - \paren {-i z}^n} {n!} }$ $\ds$ $=$ $\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} - \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }$ split into even and odd $n$ $\ds$ $=$ $\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} - \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!}$ as $\paren {-i z}^{2 n} = \paren {i z}^{2 n}$ $\ds$ $=$ $\ds \frac 1 {2 i} \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}$ as $\paren {-1}^{2 n + 1} = -1$ $\ds$ $=$ $\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n + 1} } {\paren {2 n + 1}!}$ cancel $2$ $\ds$ $=$ $\ds \frac 1 i \sum_{n \mathop = 0}^\infty \frac {i \paren {-1}^n z^{2 n + 1} } {\paren {2 n + 1}!}$ as $i^{2 n + 1} = i \paren {-1}^n$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1!} }$ cancel $i$ $\ds$ $=$ $\ds \sin z$

$\blacksquare$

## Proof 2

Recall Euler's Formula:

$\exp \paren {i z} = \cos z + i \sin z$

Then, starting from the right hand side:

 $\ds \frac {\exp \paren {i z} - \exp \paren {-i z} }{2 i}$ $=$ $\ds \frac {\paren {\cos z + i \sin z} - \paren {\cos \paren {-z} + i \sin \paren {-z} } } {2 i}$ $\ds$ $=$ $\ds \frac {\paren {\cos z + i \sin z - \cos z - i \sin \paren {-z} } } {2 i}$ Cosine Function is Even $\ds$ $=$ $\ds \frac {i \sin z - i \sin \paren {-z} } {2 i}$ $\ds$ $=$ $\ds \frac {i \sin z - i \paren {-\sin \paren {-z} } } {2 i}$ Sine Function is Odd $\ds$ $=$ $\ds \frac {2 i \sin z} {2 i}$ $\ds$ $=$ $\ds \sin z$

$\blacksquare$

## Proof 3

 $\text {(1)}: \quad$ $\ds \exp \paren {i z}$ $=$ $\ds \cos z + i \sin z$ Euler's Formula $\text {(2)}: \quad$ $\ds \exp \paren {-i z}$ $=$ $\ds \cos z - i \sin z$ Euler's Formula: Corollary $\ds \leadsto \ \$ $\ds \exp \paren {i z} - \exp \paren {-i z}$ $=$ $\ds \paren {\cos z + i \sin z} - \paren {\cos z - i \sin z}$ $(1) - (2)$ $\ds$ $=$ $\ds 2 i \sin z$ simplifying $\ds \leadsto \ \$ $\ds \frac {\exp \paren {i z} - \exp \paren {-i z} } {2 i}$ $=$ $\ds \sin z$

$\blacksquare$

## Also presented as

This result can also be presented as:

$\sin z = \dfrac 1 2 i \paren {e^{-i z} - e^{i z} }$