જો a = frac{1 - i sqrt{3}}{2} હોય, તો List-I | vedclass

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(B) આપેલ છે કે $a = \frac{1 - i \sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.તેથી $\bar{a} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$(i)$ $a \bar{a

Vedclass · Accepted Answer

$D, A, B, F$

Vedclass · Answer

(B) આપેલ છે કે $a = \frac{1 - i \sqrt{3}}{2} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$.તેથી $\bar{a} = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.$(i)$ $a \bar{a} = |a|^2 = \left(\frac{1}{2}ight)^2 + \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{1}{4} + \frac{3}{4} = 1$. જે $(D)$ સાથે જોડાય છે.$(ii)$ $\arg \left(\frac{1}{\bar{a}}ight) = \arg(a) = 	an^{-1}\left(\frac{-\sqrt{3}/2}{1/2}ight) = 	an^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$. જે $(A)$ સાથે જોડાય છે.$(iii)$ $a - \bar{a} = \left(\frac{1}{2} - i \frac{\sqrt{3}}{2}ight) - \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}ight) = -i \sqrt{3}$. જે $(B)$ સાથે જોડાય છે.$(iv)$ $\frac{4}{3a} = \frac{4}{3} \cdot \frac{1}{a} = \frac{4}{3} \cdot \frac{\bar{a}}{|a|^2} = \frac{4}{3} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}ight) = \frac{2}{3} + i \frac{2\sqrt{3}}{3} = \frac{2}{3} + i \frac{2}{\sqrt{3}}$.આમ, $\operatorname{Im}\left(\frac{4}{3a}ight) = \frac{2}{\sqrt{3}}$. જે $(F)$ સાથે જોડાય છે.તેથી, સાચું જોડાણ $(i)-D, (ii)-A, (iii)-B, (iv)-F$ છે.

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

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