Answer
$3\;[\;cos(\pi+2n\pi )+isin(\pi+2n\pi )\;]\;=3\;e^{i(\pi+2n\pi) }\\\\$
Work Step by Step
remember;
$R=|z|=\sqrt{a^2+b^2} \;\;\;\;\;\;and\;\;\;\;\;\;\;tan\Theta =\frac{a}{b}\\\\$
$R=|z|=\sqrt{(-3)^2}\;=\;\sqrt{9}=3\\\\$
$tan\Theta =\frac{b}{a}\;=\;\frac{0}{-3}=0\\\\$
$\Theta =tan^{-1}(0) =\pi\\\\$
$R(cos\Theta +isin\Theta )= Re^{i\Theta }\\\\$
$\Rightarrow 3\;[\;cos(\pi)+isin(\pi)\;]\;=2\;e^{i\pi}\\\\$
Notice that because of periodicity it is the same number if we replace $\Theta$ with $\Theta +2n\pi$ for any integer $n$.
$3\;[\;cos(\pi+2n\pi )+isin(\pi+2n\pi )\;]\;=3\;e^{i(\pi+2n\pi) }\\\\$
