Answer
$2\;[\;cos(\frac{2\pi}{3}+2n\pi )+isin(\frac{2\pi}{3}+2n\pi )\;]\;=2\;e^{i(\frac{2\pi}{3}+2n\pi) }\\\\$
Work Step by Step
remember;
$R=|z|=\sqrt{a^2+b^2} \;\;\;\;\;\;and\;\;\;\;\;\;\;tan\Theta =\frac{a}{b}\\\\$
$R=|z|=\sqrt{(-1)^2+(\sqrt{3})^2}\;=\;\sqrt{4}=2\\\\$
$tan\Theta =\frac{b}{a}\;=\;\frac{\sqrt{3}}{-1}=-\sqrt{3}\\\\$
$\Theta =tan^{-1}(-\sqrt{3}) =\frac{2\pi}{3} \\\\$
$R(cos\Theta +isin\Theta )= Re^{i\Theta }\\\\$
$\Rightarrow 2\;[\;cos(\frac{2\pi}{3})+isin(\frac{2\pi}{3})\;]\;=2\;e^{i\frac{2\pi}{3}}\\\\$
Notice that because of periodicity it is the same number if we replace $\Theta$ with $\Theta +2n\pi$ for any integer $n$.
$2\;[\;cos(\frac{2\pi}{3}+2n\pi )+isin(\frac{2\pi}{3}+2n\pi )\;]\;=2\;e^{i(\frac{2\pi}{3}+2n\pi) }\\\\$
