The complex exponential is 2πi-periodic

Statement

For each $z \in \mathbb{C}$, $$e^{z + 2 \pi i} = e^{z},$$ where $e^z$ is the complex exponential of $z$.

Remarks

Proof: by Euler's formula, $e^{2 \pi i} = \cos 2\pi + i \sin 2\pi = 1$. By the product of exponentials, $e^{z + 2\pi i} = e^z e^{2\pi i} = e^z$. The periods of $\exp$ are exactly the numbers $2 \pi i k$ with $k \in \mathbb{Z}$, by exp z = 1 iff z is in 2πiZ. So $\exp$ is not injective on $\mathbb{C}$. It is injective on each horizontal strip $\{z : a < \operatorname{Im} z \le a + 2\pi\}$ with $a \in \mathbb{R}$.

Explanations

L1 human

Adding $2 \pi i$ to $z$ adds a full turn to the direction of $e^z$ and does not change its size. So $e^z$ takes the same value again: the exponential repeats itself along each vertical line, with step $2\pi$.