# How do you find the square root of 16(cos60^@+isin60^@)?

Dec 22, 2016

$2 \sqrt{3} + 2 i$

#### Explanation:

$z = 16 \left(\cos {60}^{0} + i \sin {60}^{0}\right)$

sqrtz=z^(1/2)=sqrt(16(cos60^0+isin60^0)

${z}^{\frac{1}{2}} = 4 {\left(\cos 60 + i \sin 60\right)}^{\frac{1}{2}}$

now using DeMoivre's theorem

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta , \forall n \in \mathbb{R}$

${z}^{\frac{1}{2}} = 4 \left(\cos \left(\frac{60}{2}\right) + i \sin \left(\frac{60}{2}\right)\right)$

${z}^{\frac{1}{2}} = 4 \left(\cos {30}^{0} + i \sin {30}^{0}\right)$

${z}^{\frac{1}{2}} = 4 \left(\frac{\sqrt{3}}{2} + i \frac{1}{2}\right)$

${z}^{\frac{1}{2}} = 2 \sqrt{3} + 2 i$