Answer
$\;[\;cos(\frac{3\pi}{2}+2n\pi )+isin(\frac{3\pi}{2}+2n\pi )\;]\;=\;e^{i(\frac{3\pi}{2}+2n\pi) }\\\\$
Work Step by Step
remember;
$R=|z|=\sqrt{a^2+b^2} \;\;\;\;\;\;and\;\;\;\;\;\;\;tan\Theta =\frac{a}{b}\\\\$
$R=|z|=\sqrt{(-1)^2}\;=\;\sqrt{1}=1\\\\$
$tan\Theta =\frac{-1}{0}\;=\;undefined\\\\$
$\Theta =\frac{3\pi}{2}\\\\$
$R(cos\Theta +isin\Theta )= Re^{i\Theta }\\\\$
$\Rightarrow 1\;[\;cos(\frac{3\pi}{2})+isin(\frac{3\pi}{2})\;]\;=1\;e^{i\frac{3\pi}{2}}\\\\$
Notice that because of periodicity it is the same number if we replace $\Theta$ with $\Theta +2n\pi$ for any integer $n$.
$\;[\;cos(\frac{3\pi}{2}+2n\pi )+isin(\frac{3\pi}{2}+2n\pi )\;]\;=\;e^{i(\frac{3\pi}{2}+2n\pi) }\\\\$
