If left| z - frac{1 + 3i}{2} right| = frac{sq | vedclass

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(C) Let the complex numbers be $z_1 = z$,$z_2 = z e^{i \pi / 3}$,and $z_3 = z(1 + e^{i \pi / 3})$.$PQ = |z_2 - z_1| = |z e^{i \pi / 3} - z| = |z| |e^{

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$\frac{\sqrt{3}}{4} |z|^2$

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(C) Let the complex numbers be $z_1 = z$,$z_2 = z e^{i \pi / 3}$,and $z_3 = z(1 + e^{i \pi / 3})$.$PQ = |z_2 - z_1| = |z e^{i \pi / 3} - z| = |z| |e^{i \pi / 3} - 1|$.Using $e^{i 	heta} = \cos 	heta + i \sin 	heta$,we have $|e^{i \pi / 3} - 1| = |(\cos \frac{\pi}{3} - 1) + i \sin \frac{\pi}{3}| = |-\frac{1}{2} + i \frac{\sqrt{3}}{2}| = \sqrt{(-\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$.Thus,$PQ = |z|$.$QR = |z_3 - z_2| = |z(1 + e^{i \pi / 3}) - z e^{i \pi / 3}| = |z|$.$PR = |z_3 - z_1| = |z(1 + e^{i \pi / 3}) - z| = |z e^{i \pi / 3}| = |z|$.Since $PQ = QR = PR = |z|$,the triangle $PQR$ is an equilateral triangle with side length $s = |z|$.The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} s^2$.Therefore,the area of $	riangle PQR = \frac{\sqrt{3}}{4} |z|^2$.

If $\left| z - \frac{1 + 3i}{2} \right| = \frac{\sqrt{10}}{2}$ and $P$,$Q$,and $R$ are points representing the complex numbers $z$,$z e^{i \pi / 3}$,and $z(1 + e^{i \pi / 3})$ respectively in the Argand plane,then the area of the triangle $PQR$ is:

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