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Published December 21, 2020

Euler's formula states that for any real number $$x$$:

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

When $$x = \pi$$, Euler's formula evaluates to:

$e^{i\pi} + 1 = 0$

This is known as Euler's identity and it links five fundamental mathematical constants ($$\pi$$, $$e$$, $$i$$, $$0$$, and $$1$$) in a beautifully simple equation. Richard Feynman called the equation "the most remarkable formula in mathematics."

The formula can be interpreted as saying that the function $$e^{i\phi}$$ is a unit complex number, i.e. it traces out the unit circle in the complex plane as $$\phi$$ ranges through the real numbers.

We can derive Euler's formula from the Maclaurin series expansions of $$\sin x$$, $$\cos x$$, and $$e^x$$:

\begin{align} \sin x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} \\ &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\ \cos x &= \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n} \\ &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \\ e^x &= \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align}

Thus:

\begin{align} e^{ix} &= \sum_{n = 0}^{\infty} \frac{(ix)^n}{n!} \\ e^{ix} &= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \cdots \\ &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \cdots \\ &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \cdots \right) + \left( ix - \frac{ix^3}{3!} + \frac{ix^5}{5!} - \frac{ix^7}{7!} + \cdots \right) \\ &= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\ &= \cos x + i\sin x \end{align}

This yields Euler's identity. Substitute $$\pi$$ for $$x$$. Since $$\cos \pi = -1$$ and $$\sin \pi = 0$$, it follows that:

$e^{i\pi} = -1 + 0i$

or

$e^{i\pi} + 1 = 0$