Question Number 581 by 123456 last updated on 31/Jan/15
![z^2 =cos θ+isin θ z=?](https://www.tinkutara.com/question/Q581.png)
$${z}^{\mathrm{2}} =\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta \\ $$$${z}=? \\ $$
Answered by ssahoo last updated on 31/Jan/15
![e^(iθ) =cos θ +isin θ=z^2 z=e^((iθ)/2) =cos (θ/2)+isin (θ/2)](https://www.tinkutara.com/question/Q583.png)
$${e}^{{i}\theta} =\mathrm{cos}\:\theta\:+{i}\mathrm{sin}\:\theta={z}^{\mathrm{2}} \\ $$$${z}={e}^{\frac{{i}\theta}{\mathrm{2}}} =\mathrm{cos}\:\frac{\theta}{\mathrm{2}}+{i}\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$