Q47 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
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$2\left[\cos \left(\frac{-4 \pi}{3}\right)+i \sin \left(\frac{-4 \pi}{3}\right)\right]$
- Correct answerB$2\left[\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(\frac{-5 \pi}{6}\right)\right]$
- C$2\left[\cos \left(\frac{-\pi}{6}\right)+i \sin \left(\frac{-\pi}{6}\right)\right]$
- D$2\left[\cos \left(\frac{-2 \pi}{3}\right)+i \sin \left(\frac{-2 \pi}{3}\right)\right]$
Solution
We have,
$$
\begin{aligned}
& z=\sqrt{2} \sqrt{1+\sqrt{3} i} \Rightarrow z=\sqrt{2+2 \sqrt{3}} i \\
& z=\sqrt{(\sqrt{3}+i)^2} \quad \Rightarrow z= \pm(\sqrt{3}+i)
\end{aligned}
$$
$z$ lie in third quadrant
$$
\begin{aligned}
& \therefore \quad z=-\sqrt{3}-i \\
& \Rightarrow|z|=\sqrt{(-\sqrt{3})^2+(-1)^2}=\sqrt{3+1}=\sqrt{4}=2 \\
& \text { and } \tan \theta=\left|\frac{-1}{-\sqrt{3}}\right|=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6}
\end{aligned}
$$
Since $\theta$ lie in 3rd quadrant
$$
\begin{aligned}
& \therefore \arg (z)=-\left(\pi-\frac{\pi}{6}\right)=-\frac{5 \pi}{6} \\
& \text { Hence } z=2\left(\cos \left(\frac{-5 \pi}{6}\right)+i \sin \left(\frac{-5 \pi}{6}\right)\right)
\end{aligned}
$$
Hence, the correct answer is option: 'B'.