Euler's Identity

Theorem

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

Proof
Follows directly from Euler's Formula $e^{i \theta} = \cos \theta + i \sin \theta$, by plugging in $\theta = \pi$:


 * $e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + i \times 0 + 1 = 0$

Also presented as
This result can also be presented as:


 * $e^{i \pi} = -1$