In the Argand plane,no value of sqrt[3]{1-i s | vedclass

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(A) Let $z = 1 - i \sqrt{3}$.In polar form,$z = r(\cos 	heta + i \sin 	heta)$,where $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.The angle $	h

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First quadrant

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(A) Let $z = 1 - i \sqrt{3}$.In polar form,$z = r(\cos 	heta + i \sin 	heta)$,where $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$.The angle $	heta$ is given by $	an 	heta = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since $z$ is in the fourth quadrant,$	heta = -\frac{\pi}{3}$.Thus,$z = 2(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.The cube roots are given by $w_k = \sqrt[3]{2} \left( \cos \left( \frac{-\pi/3 + 2k\pi}{3} ight) + i \sin \left( \frac{-\pi/3 + 2k\pi}{3} ight) ight)$ for $k = 0, 1, 2$.For $k=0$: $	heta_0 = -\frac{\pi}{9} = -20^\circ$ (Fourth quadrant).For $k=1$: $	heta_1 = \frac{-\pi/3 + 2\pi}{3} = \frac{5\pi}{9} = 100^\circ$ (Second quadrant).For $k=2$: $	heta_2 = \frac{-\pi/3 + 4\pi}{3} = \frac{11\pi}{9} = 220^\circ$ (Third quadrant).None of the roots lie in the first quadrant.

In the Argand plane,no value of $\sqrt[3]{1-i \sqrt{3}}$ lies in which quadrant?

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