## Euler's Forumla

$e^{ix} = \cos x + i * \sin x$

## Proof

Using Taylor Series:

$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$

$e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \cdots$

Expanding the complex terms yield:

$e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{(ix)^3}{3!} + \frac{x^4}{4!} + \frac{(ix)^5}{5!} - \cdots$

Rearranging the terms into even and odd exponents and factoring out the imaginary number:

$e^{ix} = (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots)$

Which is equivalent to:

$e^{ix} = \cos x + i\sin x$

## Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva