Question

If $z=\sqrt{2} \sqrt{1+\sqrt{3 i}}$ represents a point $P$ in the argand plane and $P$ lies in the third quadrant, then the polar form of $z$ is

• A. $2\left[\cos \left(\frac{-4 \pi}{3}\right)+i \sin \left(\frac{-4 \pi}{3}\right)\right]$
• B. $2\left[\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(\frac{-5 \pi}{6}\right)\right]$
• C. $2\left[\cos \left(\frac{-\pi}{6}\right)+i \sin \left(\frac{-\pi}{6}\right)\right]$
• D. $2\left[\cos \left(\frac{-2 \pi}{3}\right)+i \sin \left(\frac{-2 \pi}{3}\right)\right]$

Answer 1

Answer: (B)

We have,
$z=\sqrt{2} \sqrt{1+\sqrt{3} i} \Rightarrow z=\sqrt{2+2 \sqrt{3}} i $
$z=\sqrt{(\sqrt{3}+i)^{2}} \Rightarrow z=\pm(\sqrt{3}+i)$
$z$ lie in third quadrant
$\therefore \quad z=-\sqrt{3}-i$
$\Rightarrow |z|=\sqrt{(-\sqrt{3})^{2}+(-1)^{2}}=\sqrt{3+1}=\sqrt{4}=2$
and $\tan \theta=\left|\frac{-1}{-\sqrt{3}}\right|=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}$
Since $\theta$ lie in 3 rd quadrant
$\therefore \arg (z)=-\left(\pi-\frac{\pi}{6}\right)=-\frac{5 \pi}{6}$
Hence $z=2\left(\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(\frac{-5 \pi}{6}\right)\right)$