# Cosine Exponential Formulation

## Theorem

For any complex number $z \in \C$:

$\cos z = \dfrac {\map \exp {i z} + \map \exp {-i z} } 2$

where:

$\exp z$ denotes the exponential function
$\cos z$ denotes the complex cosine function
$i$ denotes the inaginary unit.

### Real Domain

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

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

## Proof 1

Recall the definition of the cosine function:

 $\ds \cos z$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$ $\ds$ $=$ $\ds 1 - \frac {z^2} {2!} + \frac {z^4} {4!} - \frac {z^6} {6!} + \cdots + \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!} + \cdots$

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:

 $\ds \frac {\exp \paren {i z} + \exp \paren {-i z} } 2$ $=$ $\ds \frac 1 2 \paren {\sum_{n \mathop = 0}^\infty \frac {\paren {i z}^n} {n!} + \sum_{n \mathop = 0}^\infty \frac {\paren {-i z}^n} {n!} }$ $\ds$ $=$ $\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^n + \paren {-i z}^n} {n!} }$ Cosine Function is Absolutely Convergent $\ds$ $=$ $\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \paren {\frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!} + \frac {\paren {i z}^{2 n + 1} + \paren {-i z}^{2 n + 1} } {\paren {2 n + 1}!} }$ split into even and odd $n$ $\ds$ $=$ $\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} + \paren {-i z}^{2 n} } {\paren {2 n}!}$ $\paren {-i z}^{2 n + 1} = -\paren {i z}^{2 n + 1}$ $\ds$ $=$ $\ds \frac 1 2 \sum_{n \mathop = 0}^\infty \frac {2 \paren {i z}^{2 n} } {\paren {2 n}!}$ $\left({ -1 }\right)^{2n} = 1$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {i z}^{2 n} } {\paren {2 n}!}$ cancel $2$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n} } {\paren {2 n}!}$ $i^{2 n} = \paren {-1}^n$ $\ds$ $=$ $\ds \cos z$

$\blacksquare$

## Proof 2

Recall Euler's Formula:

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

Then, starting from the right hand side:

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

$\blacksquare$

## 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 \ \$ $\ds \exp \paren {i z} + \exp \paren {-i z}$ $=$ $\ds \paren {\cos z + i \sin z} + \paren {\cos z - i \sin z}$ $(1) + (2)$ $\ds$ $=$ $\ds 2 \cos z$ simplifying $\ds \leadsto \ \$ $\ds \frac {\exp \paren {i z} + \exp \paren {-i z} } 2$ $=$ $\ds \cos z$

$\blacksquare$

## Also presented as

This result can also be presented as:

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