For the complex exponential functions holds

$$e^{z+2\,\pi\,\ii}=e^z$$ The periodicity of the cosine follows from the following calculation: $$\begin{aligned}\cos(z+2\pi)&=\frac{1}{2}\left(e^{(z+2\pi)\cdot\ii}+e^{-(z+2\pi)\cdot\ii}\right)\\ &\phantom{rstuvwxyz}\color{blue}{\text{definition}}\\ &=\frac{1}{2}\left(\e^{z\cdot\ii}+\e^{-z\cdot\ii}\right) \\ &\phantom{rstuvwxyz}\color{blue}{\text{periodicity of }\exp}\\ &=\cos(z) \\ &\phantom{rstuvwxyz} \color{blue}{\text{definition}}\end{aligned}$$ The proof that the sine is periodic with period $2\pi$ is the same.

For the maths enthusiast, we give the proof of the theorem that

$$\cos^2(z)+\sin^2(z)=1\qquad\text{for any complex number }z$$ This proof is in fact no more than the use of the definitions of trigonometric functions and expansion of the squares. $$\begin{aligned}\cos^2(z)+\sin^2(z) &= \left(\frac{e^{z\,\ii}+e^{-z\,\ii}}{2}\right)^2+ \left(\frac{e^{z\,\ii}-e^{-z\,\ii}}{2\ii}\right)^2\\ & \phantom{uvwxyz}\color{blue}{\text{definitions}} \\ &= \frac{\bigl(e^{z\,\ii}\bigr)^2+2\cdot e^{z\,\ii}\cdot e^{-z\,\ii}+\bigl(e^{-z\,\ii}\bigr)^2}{4}\\ &\phantom{=}\quad {}+ \frac{\bigl(e^{z\,\ii}\bigr)^2-2\cdot e^{z\,\ii}\cdot e^{-z\,\ii}+\bigl(e^{-z\,\ii}\bigr)^2}{4\ii^2}\\ & \phantom{uvwxyz}\color{blue}{\text{sum and difference formula of squares}} \\ &= \frac{e^{2z\,\ii}+2+e^{-2z\,\ii}}{4} - \frac{e^{2z\,\ii} - 2+e^{-2z\,\ii}}{4}\\ & \phantom{uvwxyz}\color{blue}{\text{calculation rules for powers and }\ii^2=-1} \\ &= \frac{e^{2z\,\ii}+2+e^{-2z\,\ii}-e^{2z\,\ii} + 2-e^{-2z\,\ii}}{4}\\ & \phantom{uvwxyz}\color{blue}{\text{everything under one common denominator}}\\ &= \frac{4}{4}= 1 \\ & \phantom{uvwxyz}\color{blue}{\text{simplification}}\end{aligned}$$