---
tag: D1V15Q69Q8
kind: proof
name: Proof that the complex exponential is 2πi-periodic
msc: [33B10]
proves: 02P2PYH3X3
uses: [054VXHMCA3, 04R3CY41VJ, 08W5RN375X, 02KTRHQMHN, 0P4NQD746K, 0WM7DJCM6G]
sources:
  - url: https://en.wikipedia.org/wiki/Exponential_function
    locator: Complex exponential
---

## Proof

We compute $e^{2 \pi i}$ and use the product rule for exponentials.

1. By [[Euler's formula|04R3CY41VJ]] with $z = 2\pi$, we get $e^{2 \pi i} = \cos 2\pi + i \sin 2\pi$. Here $e^w$ is the [[complex exponential|054VXHMCA3]], and $\cos$ and $\sin$ are given by their power series.
2. At a real argument, these series are the real [[cosine|08W5RN375X]] and the real [[sine|02KTRHQMHN]]. So $\cos 2\pi$ and $\sin 2\pi$ are the values of the real functions.
3. Only the constant terms of the series are nonzero at $0$. So $\cos 0 = 1$ and $\sin 0 = 0$.
4. Since [[sine and cosine are $2\pi$-periodic|0P4NQD746K]], $\cos 2\pi = \cos 0 = 1$ and $\sin 2\pi = \sin 0 = 0$.
5. By steps 1 and 4, $e^{2 \pi i} = 1$.
6. Let $z \in \C$. By the [[product of exponentials|0WM7DJCM6G]], $e^{z + 2\pi i} = e^z e^{2 \pi i} = e^z \cdot 1 = e^z$.

## Idea

The exponential turns sums into products, so a shift by $2 \pi i$ multiplies $e^z$ by $e^{2\pi i}$. Euler's formula identifies $e^{2 \pi i}$ with the point at angle $2\pi$ on the unit circle, which is $1$.