If a = frac{1-i sqrt{3}}{2},then the correct | vedclass

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(B) Given,$a = \frac{1-i \sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.Then,$\bar{a} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$(i) \ a \bar{a} = |a|

Vedclass · Accepted Answer

$D, A, B, F$

Vedclass · Answer

(B) Given,$a = \frac{1-i \sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.Then,$\bar{a} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$(i) \ a \bar{a} = |a|^2 = \left(\frac{1}{2}ight)^2 + \left(-\frac{\sqrt{3}}{2}ight)^2 = \frac{1}{4} + \frac{3}{4} = 1$. Matches $(D)$.$(ii) \ \arg \left(\frac{1}{\bar{a}}ight) = \arg(a) = 	an^{-1}\left(\frac{-\sqrt{3}/2}{1/2}ight) = 	an^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$. Matches $(A)$.$(iii) \ a - \bar{a} = \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}ight) - \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}ight) = -i \sqrt{3}$. Matches $(B)$.$(iv) \ \frac{4}{3a} = \frac{4}{3} \cdot \frac{1}{a} = \frac{4}{3} \cdot \frac{\bar{a}}{|a|^2} = \frac{4}{3} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}ight) = \frac{2}{3} + i \frac{2\sqrt{3}}{3} = \frac{2}{3} + i \frac{2}{\sqrt{3}}$.Thus,$\operatorname{Im}\left(\frac{4}{3a}ight) = \frac{2}{\sqrt{3}}$. Matches $(F)$.The correct sequence is $(i)-D, (ii)-A, (iii)-B, (iv)-F$.

List-$I$	List-$II$
$(i) \ a \bar{a}$	$(A) -\frac{\pi}{3}$
$(ii) \ \arg \left(\frac{1}{\bar{a}}\right)$	$(B) -i \sqrt{3}$
$(iii) \ a-\bar{a}$	$(C) \frac{2i}{\sqrt{3}}$
$(iv) \ \operatorname{Im}\left(\frac{4}{3a}\right)$	$(D) 1$
	$(E) \frac{\pi}{3}$
	$(F) \frac{2}{\sqrt{3}}$

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