Find all $z$ such that $e^{2\pi i z}=1$

We have that for $z=x+iy$

$$1=|1|=|e^{2\pi i(x+iy)}|=|e^{2\pi i x}e^{-2\pi y}|=e^{-2\pi y}$$

Since $-2\pi y$ is a real number, we can take logarithms to get

$$0=\ln(1)=\ln(e^{-2\pi y})=-2\pi y$$

$$\Rightarrow 0=y$$

Alright, so we know $z=x$ for some real $x$. But then

$$1=e^{2\pi i x}=\cos(2\pi x)+i\sin(2\pi x)$$

Since $1$ is purely real, this implies

$$0=\sin(2\pi x)$$

$$\Rightarrow 2\pi x=\pi k\text{ for some }k\in\mathbb{Z}$$

$$\Rightarrow x=\frac{k}{2}\text{ for some }k\in\mathbb{Z}$$

If we use this information with the real part of the expression above, we get

$$1=\cos(2\pi x)=\cos(\pi k)\text{ for some }k\in\mathbb{Z}$$

$$\Rightarrow \pi k=2\pi m\text{ for some }m\in\mathbb{Z}$$

$$\Rightarrow k=2 m\text{ for some }m\in\mathbb{Z}$$

$$\Rightarrow x=\frac{k}{2}=\frac{2m}{2}=m\text{ for some }m\in\mathbb{Z}$$

We conclude $x$ must necessarily be an integer. Since it is easy to check that any integer is indeed a solution, we conclude the solution set is the integers.