Two values of (-8-8 sqrt{3} i)^{1/4} are | vedclass

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(A) Let $z = -8-8 \sqrt{3} i$. We can write this in polar form as $z = r(\cos 	heta + i \sin 	heta)$.Here,$r = \sqrt{(-8)^2 + (-8 \sqrt{3})^2} = \sq

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$\sqrt{3}-i, -1-\sqrt{3} i$

Vedclass · Answer

(A) Let $z = -8-8 \sqrt{3} i$. We can write this in polar form as $z = r(\cos 	heta + i \sin 	heta)$.Here,$r = \sqrt{(-8)^2 + (-8 \sqrt{3})^2} = \sqrt{64 + 192} = \sqrt{256} = 16$.Since both real and imaginary parts are negative,the angle $	heta$ lies in the third quadrant.$	an 	heta = \frac{-8 \sqrt{3}}{-8} = \sqrt{3}$,so $	heta = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$.Thus,$z = 16(\cos(\frac{4 \pi}{3} + 2k \pi) + i \sin(\frac{4 \pi}{3} + 2k \pi))$ for $k = 0, 1, 2, 3$.The roots are given by $z^{1/4} = 16^{1/4}(\cos(\frac{4 \pi/3 + 2k \pi}{4}) + i \sin(\frac{4 \pi/3 + 2k \pi}{4})) = 2(\cos(\frac{\pi}{3} + \frac{k \pi}{2}) + i \sin(\frac{\pi}{3} + \frac{k \pi}{2}))$.For $k=0$: $2(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) = 2(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = 1 + i \sqrt{3}$.For $k=1$: $2(\cos \frac{5 \pi}{6} + i \sin \frac{5 \pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$.For $k=2$: $2(\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) = 2(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -1 - i \sqrt{3}$.For $k=3$: $2(\cos \frac{11 \pi}{6} + i \sin \frac{11 \pi}{6}) = 2(\frac{\sqrt{3}}{2} - i \frac{1}{2}) = \sqrt{3} - i$.The values are $\sqrt{3}-i$ and $-\sqrt{3}+i$ (among others). Comparing with options,option $C$ contains $-\sqrt{3}+i$ and $\sqrt{3}+i$ is not a root,but checking the provided options,the correct set is $A$.

Two values of $(-8-8 \sqrt{3} i)^{1/4}$ are

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