Question

If $z=\sqrt{2}\sqrt{1+\sqrt{3i}}$ 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
$\therefore z=-\sqrt{3}-i$
$\Rightarrow |z|=\sqrt{(-\sqrt{3}{)}^2+(-1{)}^2}=\sqrt{3+1}=\sqrt{4}=2$
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 $z=2\left(\cos \left(\frac{-5\pi }{6}\right)+i\sin \left(\frac{-5\pi }{6}\right)\right)$