Mathematics | TS EAMCET 2019 | Discussion Page

If $z=\sqrt{2}\sqrt{1+\sqrt{3i}}$ repesents 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) $2\left[ \cos \left(\frac{- 2\pi}{3}\right)+i \sin \left(\frac{- 2\pi}{3}\right)\right]$

Option B

We have

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

$z=\sqrt{(\sqrt{3}+i)^{2}}\Rightarrow z=\pm (\sqrt{3}+i)$

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

$\tan\theta=|\frac{-1}{-\sqrt{3}}|=\frac{1}{\sqrt{3}}\Rightarrow\theta=\frac{\pi}{6}$

Since $\theta$ lie in 3rd quadrant

$\therefore$ $arg(z)=-\left(\pi-\frac{\pi}{6}\right)= -\frac{5 \pi}{6}$

Hence, $z=2\left( \cos \left( -\frac{5 \pi}{6}\right) +i \sin \left(-\frac{5 \pi}{6}\right)\right)$

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