Euler's Identity
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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
Euler's Identity 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
- For a video presentation of the contents of this page, visit the Khan Academy.