If z = sqrt{2} sqrt{1 + sqrt{3} i} represents | vedclass

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(B) Given $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$.We can write $1 + \sqrt{3} i = \frac{1}{2} (2 + 2 \sqrt{3} i) = \frac{1}{2} ((\sqrt{3})^2 + i^2 + 2 \sq

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$2 \left[ \cos \left( \frac{-5 \pi}{6} \right) + i \sin \left( \frac{-5 \pi}{6} \right) \right]$

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(B) Given $z = \sqrt{2} \sqrt{1 + \sqrt{3} i}$.We can write $1 + \sqrt{3} i = \frac{1}{2} (2 + 2 \sqrt{3} i) = \frac{1}{2} ((\sqrt{3})^2 + i^2 + 2 \sqrt{3} i) = \frac{1}{2} (\sqrt{3} + i)^2$.Thus,$z = \sqrt{2} \cdot \frac{1}{\sqrt{2}} (\sqrt{3} + i) = \pm (\sqrt{3} + i)$.Since $z$ lies in the third quadrant,both real and imaginary parts must be negative.Therefore,$z = -\sqrt{3} - i$.The modulus is $|z| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$.The argument $	heta$ in the third quadrant is given by $	heta = -(\pi - 	an^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5 \pi}{6}$.Thus,the polar form is $z = 2 \left[ \cos \left( \frac{-5 \pi}{6} ight) + i \sin \left( \frac{-5 \pi}{6} ight) ight]$.

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

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