Answer
$2\;[\;cos(\frac{11\pi}{6}+2n\pi )+isin(\frac{11\pi}{6}+2n\pi )\;]\;=2\;e^{i(\frac{11\pi}{6}+2n\pi) }\\\\$
Work Step by Step
remember;
$R=|z|=\sqrt{a^2+b^2} \;\;\;\;\;\;and\;\;\;\;\;\;\;tan\Theta =\frac{a}{b}\\\\$
$R=|z|=\sqrt{(\sqrt{3})^2+(-1)^2}\;=\;\sqrt{4}=2\\\\$
$tan\Theta =\frac{-1}{\sqrt{3}}\;\\\\$
$\Theta =tan^{-1}(\frac{-1}{\sqrt{3}}) =\frac{11\pi}{6}\\\\$
$R(cos\Theta +isin\Theta )= Re^{i\Theta }\\\\$
$\Rightarrow 2\;[\;cos(\frac{11\pi}{6})+isin(\frac{11\pi}{6})\;]\;=2\;e^{i\frac{11\pi}{6}}\\\\$
Notice that because of periodicity it is the same number if we replace $\Theta$ with $\Theta +2n\pi$ for any integer $n$.
$2\;[\;cos(\frac{11\pi}{6}+2n\pi )+isin(\frac{11\pi}{6}+2n\pi )\;]\;=2\;e^{i(\frac{11\pi}{6}+2n\pi) }\\\\$
