If left(frac{sqrt{3}+i}{sqrt{3}-i}right)^4+le | vedclass

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(B) Let $z = \frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $z = \frac{(\sqrt{3}+i)^2}{3+1} = \frac{3-1+

Vedclass · Accepted Answer

$\operatorname{cis}\left(\frac{3 \pi}{2}\right)$

Vedclass · Answer

(B) Let $z = \frac{\sqrt{3}+i}{\sqrt{3}-i}$. Multiplying numerator and denominator by $\sqrt{3}+i$,we get $z = \frac{(\sqrt{3}+i)^2}{3+1} = \frac{3-1+2i\sqrt{3}}{4} = \frac{2+2i\sqrt{3}}{4} = \frac{1}{2} + i\frac{\sqrt{3}}{2} = \operatorname{cis}\left(\frac{\pi}{3}ight)$.Then the given expression is $z^4 + (\bar{z})^4 = \operatorname{cis}\left(\frac{4\pi}{3}ight) + \operatorname{cis}\left(-\frac{4\pi}{3}ight) = 2 \cos\left(\frac{4\pi}{3}ight) = 2 \cos\left(\pi + \frac{\pi}{3}ight) = -2 \cos\left(\frac{\pi}{3}ight) = -2 	imes \frac{1}{2} = -1$.So,$r \operatorname{cis} 	heta = -1 = 1 \cdot \operatorname{cis}(\pi)$.We need to find $\sqrt{r \operatorname{cis} 	heta} = \sqrt{-1} = \sqrt{\operatorname{cis}(\pi)}$.The square roots of $\operatorname{cis}(\pi)$ are $\operatorname{cis}\left(\frac{\pi + 2k\pi}{2}ight)$ for $k=0, 1$.For $k=0$,we get $\operatorname{cis}\left(\frac{\pi}{2}ight) = i$.For $k=1$,we get $\operatorname{cis}\left(\frac{3\pi}{2}ight) = -i$.Comparing with the options,$\operatorname{cis}\left(\frac{3\pi}{2}ight)$ is the correct value.

If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \operatorname{cis} \theta$,then one of the values of $\sqrt{r \operatorname{cis} \theta}$ is

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