If Z_1, Z_2, Z_3 are three complex numbers wi | vedclass

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(A) Given $|Z_1| = |Z_2| = |Z_3| = 1$. We know that $|Z|^2 = Z \overline{Z}$. Given $|Z_1-Z_2|^2+|Z_1-Z_3|^2=4$. Expanding this,we get: $(Z_1-Z_2)(\ov

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(A) Given $|Z_1| = |Z_2| = |Z_3| = 1$. We know that $|Z|^2 = Z \overline{Z}$. Given $|Z_1-Z_2|^2+|Z_1-Z_3|^2=4$. Expanding this,we get: $(Z_1-Z_2)(\overline{Z_1}-\overline{Z_2}) + (Z_1-Z_3)(\overline{Z_1}-\overline{Z_3}) = 4$. $Z_1\overline{Z_1} - Z_1\overline{Z_2} - \overline{Z_1}Z_2 + Z_2\overline{Z_2} + Z_1\overline{Z_1} - Z_1\overline{Z_3} - \overline{Z_1}Z_3 + Z_3\overline{Z_3} = 4$. Since $|Z_1|^2 = |Z_2|^2 = |Z_3|^2 = 1$,we have: $1 - (Z_1\overline{Z_2} + \overline{Z_1}Z_2) + 1 + 1 - (Z_1\overline{Z_3} + \overline{Z_1}Z_3) + 1 = 4$. $4 - (Z_1\overline{Z_2} + \overline{Z_1}Z_2 + Z_1\overline{Z_3} + \overline{Z_1}Z_3) = 4$. Therefore,$Z_1\overline{Z_2} + \overline{Z_1}Z_2 + Z_1\overline{Z_3} + \overline{Z_1}Z_3 = 0$.

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

If $Z_1, Z_2, Z_3$ are three complex numbers with unit modulus such that $|Z_1-Z_2|^2+|Z_1-Z_3|^2=4$,then $Z_1 \overline{Z_2}+\overline{Z_1} Z_2+Z_1 \overline{Z_3}+\overline{Z_1} Z_3=$

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