periodicity of exponential function

Theorem.

The only periods of the complex exponential function  $z\mapsto e^z$  are the multiples of $2\pi i$.  Thus the function is one-periodic.

Proof.  Let $\omega$ be any period of the exponential function, i.e.  $e^{z+\omega }=e^z{e}^{\omega }=e^z$  for all  $z\in ℂ$.  Because $e^z$ is always $\ne 0$, we have

$e^{\omega }=\mathrm{\hspace{0.33em}1}.$(1)

If we set  $\omega =:a+ib$  with $a$ and $b$ reals, (1) gets the form

$e^a\cos b+i{e}^a\sin b=\mathrm{\hspace{0.33em}1},$(2)

which implies (see equality of complex numbers)

$$e^a\cos b=\mathrm{\hspace{0.33em}1},e^a\sin b=\mathrm{\hspace{0.33em}0}.$$

As these equations are squared and added, we obtain  $e^{2a}=1$  which , since $a$ is real, that  $a=0$.  Thus the preceding equations get the form

$$\cos b=\mathrm{\hspace{0.33em}1},\sin b=\mathrm{\hspace{0.33em}0}.$$

These result that  $b=n\cdot 2\pi$  and therefore

$$\omega =n\cdot 2\pi i\mathit{\hspace{1em}\hspace{1em}}(n=\mathrm{\hspace{0.33em}0},\pm 1,\pm 2,\pm 3,\mathrm{\dots })$$

Q.E.D.