# Square root of complex numbers?

## Find the square roots of complex numbers $z = \frac{144}{5 - 2 i}$

Feb 5, 2018

$\sqrt{\frac{144}{5 - 2 i}} = \frac{12}{\sqrt[4]{29}} \left(\cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right)\right)$

where $\theta = {\tan}^{- 1} \left(\frac{2}{5}\right)$

#### Explanation:

sqrt(144/(5-2i))=sqrt((144(5+2i))/((5-2i)(5+2i))

= $12 \sqrt{\frac{5 + 2 i}{29}} = \frac{12}{\sqrt{29}} \sqrt{5 + 2 i}$

Let $5 + 2 i = r \left(\cos \theta + i \sin \theta\right)$,

then $r = \sqrt{{5}^{2} + {2}^{2}} = \sqrt{29}$ and $\theta = {\tan}^{- 1} \left(\frac{2}{5}\right)$

and hence usin DeMoivre's theorem

$\sqrt{5 + 2 i} = \sqrt[4]{29} \left(\cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right)\right)$

Hence $\sqrt{\frac{144}{5 - 2 i}} = \frac{12}{\sqrt{29}} \sqrt[4]{29} \left(\cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right)\right)$

= $\frac{12}{\sqrt[4]{29}} \left(\cos \left(\frac{\theta}{2}\right) + i \sin \left(\frac{\theta}{2}\right)\right)$

where $\theta = {\tan}^{- 1} \left(\frac{2}{5}\right)$