Euler's Identity

Theorem

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


Proof

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

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

$\blacksquare$


Also presented as

can also be presented as:

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

or:

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


Also see


Source of Name

This entry was named for Leonhard Paul Euler.


Sources

  • 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(i)}$
  • 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3$: Appendix $\text A$: Euler
  • 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Euler's formula
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