exp z = 1 iff z is an integer multiple of 2πi

Statement

Let $z \in \mathbb{C}$. Then $e^z = 1$ if and only if $z = 2 \pi i k$ for some integer $k$. Here $e^z$ is the complex exponential of $z$.

Remarks

Proof: write $z = x + iy$ with $x, y$ real. Then $e^z = e^x (\cos y + i \sin y)$. If $e^z = 1$, then $e^x = |e^z| = 1$, so $x = 0$. Then $\cos y = 1$ and $\sin y = 0$, so $y = 2 \pi k$. The converse follows from Euler's formula. It follows that $e^z = e^w$ if and only if $z - w = 2 \pi i k$ for some $k \in \mathbb{Z}$. This is the reason why the complex logarithm has many values.

Explanations

L1 human

The value $1$ has size $1$ and direction $0$. So $e^z = 1$ needs a real part $0$ and an imaginary part that is a whole number of full turns, that is, a multiple of $2\pi$.