Answer
$\sqrt{2}\;[\;cos(\frac{\pi}{4}+2n\pi )+isin(\frac{\pi}{4}+2n\pi )\;]\;=\sqrt{2}\;e^{i(\frac{\pi}{4}+2n\pi) }\\\\$
Work Step by Step
remember;
$R=|z|=\sqrt{a^2+b^2} \;\;\;\;\;\;and\;\;\;\;\;\;\;tan\Theta =\frac{a}{b}\\\\$
$R=|z|=\sqrt{1^2+1^2}\;=\;\sqrt{2}\\\\$
$tan\Theta =\frac{a}{b}\;=\;\frac{1}{1}=1\\\\$
$\Theta =tan^{-1}(1)= 45^{\circ}=\frac{\pi}{4} \\\\$
$R(cos\Theta +isin\Theta )= Re^{i\Theta }\\\\$
$\Rightarrow \sqrt{2}\;[\;cos(\frac{\pi}{4})+isin(\frac{\pi}{4})\;]\;=\sqrt{2}\;e^{i\frac{\pi}{4}}\\\\$
Notice that because of periodicity it is the same number if we replace $\Theta$ with $\Theta +2n\pi$ for any integer $n$.
$\sqrt{2}\;[\;cos(\frac{\pi}{4}+2n\pi )+isin(\frac{\pi}{4}+2n\pi )\;]\;=\sqrt{2}\;e^{i(\frac{\pi}{4}+2n\pi) }\\\\$
