Question Number 581 by 123456 last updated on 31/Jan/15
$${z}^{\mathrm{2}} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$$${z}=? \\ $$
Answered by ssahoo last updated on 31/Jan/15
$${e}^{{i}\theta} =\mathrm{cos}\:\theta\:+{i}\mathrm{sin}\:\theta={z}^{\mathrm{2}} \\ $$$${z}={e}^{\frac{{i}\theta}{\mathrm{2}}} =\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+{i}\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$