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Euler's formula

The Euler's formula establishes the fundamental relationship between trigonometric functions and complex exponential functions. This formula states that for any real number $x$ and imaginary number $i$:

[math]e^{ix} = \cos(x) + i\sin(x)[/math]

where [math]e[/math] is the base of natural logarithms [math]e[/math] is approximately equal to 2.71828...

Prerequisite information[edit]

The proof[edit]

We begin by taking the Maclaurin series definition of $e^{ix}$ which is

[math]e^{ix} = 1 + ix + \frac{(ix)^{2}}{2!} + \frac{(ix)^{3}}{3!} + \ldots[/math]

This expression, for real $x$, then becomes because [math]i^2=\sqrt{-1}^2=-1[/math]:

[math]e^{ix} = 1 + ix - \frac{x^{2}}{2!} - \frac{ix^{3}}{3!} + \ldots[/math]

Rearranging these terms it can be seen this is equivalent to:

[math]e^{ix} = \left(1 - \frac{x^{2}}{2!} + \ldots\right)+i\left(x - \frac{x^{3}}{3!} + \ldots\right)[/math]

These two terms are equal to the Maclaurin Series definitions for $\cos(x)$ and $\sin(x)$ respectively. Therefore

[math]e^{ix} = \cos(x) + i\sin(x)[/math].

The last rearrangement is valid as both of these series are absolutely convergent.

Corollary to Euler's formula[edit]

A well-known corollary of the Euler's formula is:

[math]e^{-ix} = \cos(x) - i \sin(x)[/math]

The proof[edit]

Using the definition of the Euler's formula:

[math]e^{-xi}=\cos{(-x)}+i\sin{(-x)}[/math]

Because the cosine function is even, we get:

[math]e^{-xi}=\cos{(x)}+i\sin{(-x)}[/math]

Because the sine function is odd, we get:

[math]e^{-xi}=\cos{(x)}-i\sin{(x)}[/math]

All proofs that use the Euler's formula[edit]