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If $a=\frac{1-i \sqrt{3}}{2}$, then the correct matching of 'List-I from List-II is
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) $2 i / \sqrt{3}$
(iv) $\operatorname{Im}\left(\frac{4}{3 a}\right)$
(D) 1
(E) $\pi / 3$
(F) $\frac{2}{\sqrt{3}}$
correct match is
(i)
(ii)
(iii)
(iv)
Options:
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) $2 i / \sqrt{3}$
(iv) $\operatorname{Im}\left(\frac{4}{3 a}\right)$
(D) 1
(E) $\pi / 3$
(F) $\frac{2}{\sqrt{3}}$
correct match is
(i)
(ii)
(iii)
(iv)
Solution:
2683 Upvotes
Verified Answer
The correct answer is:
$\begin{array}{llll}\mathrm{D} & \mathrm{A} & \mathrm{B} & \mathrm{F}\end{array}$
Given,
$a=\frac{1-i \sqrt{3}}{2}=\frac{1}{2}-\frac{i \sqrt{3}}{2}$
$\therefore \quad \bar{a}=\frac{1}{2}+\frac{i \sqrt{3}}{2}$
(i)
$\begin{aligned} a \bar{a} & =\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) \\ & =\left(\frac{1}{2}\right)^2-i^2\left(\frac{\sqrt{3}}{2}\right)^2 \\ & =\frac{1}{4}+\frac{3}{4}=1\end{aligned}$
(ii) $\arg \left(\frac{1}{\bar{a}}\right)=\tan ^{-1}\left(\frac{-\sqrt{3}}{2} \times \frac{2}{1}\right)=-\frac{\pi}{3}$
(iii) $a-\bar{a}=\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)-\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=-i \sqrt{3}$
$a=\frac{1-i \sqrt{3}}{2}=\frac{1}{2}-\frac{i \sqrt{3}}{2}$
$\therefore \quad \bar{a}=\frac{1}{2}+\frac{i \sqrt{3}}{2}$
(i)
$\begin{aligned} a \bar{a} & =\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right) \\ & =\left(\frac{1}{2}\right)^2-i^2\left(\frac{\sqrt{3}}{2}\right)^2 \\ & =\frac{1}{4}+\frac{3}{4}=1\end{aligned}$
(ii) $\arg \left(\frac{1}{\bar{a}}\right)=\tan ^{-1}\left(\frac{-\sqrt{3}}{2} \times \frac{2}{1}\right)=-\frac{\pi}{3}$
(iii) $a-\bar{a}=\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)-\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)=-i \sqrt{3}$
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