Let {z_1}, {z_2}, {z_3} be three vertices of | vedclass

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(D) The vertices ${z_1}, {z_2}, {z_3}$ form an equilateral triangle circumscribing the circle $|z| = \frac{1}{2}$.Since the circle is the incircle of

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(D) The vertices ${z_1}, {z_2}, {z_3}$ form an equilateral triangle circumscribing the circle $|z| = \frac{1}{2}$.Since the circle is the incircle of the equilateral triangle,its center is at the origin $0$.The distance from the origin to each vertex is $R = \frac{r}{\cos(30^{\circ})} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$.However,given ${z_1} = \frac{1}{2} + \frac{\sqrt{3}i}{2}$,we note $|z_1| = \sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} = 1$.For an equilateral triangle centered at the origin,the vertices are obtained by rotating ${z_1}$ by $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ radians.${z_2} = {z_1} e^{i(2\pi/3)} = (\frac{1}{2} + \frac{\sqrt{3}}{2}i) (\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})$${z_2} = (\frac{1}{2} + \frac{\sqrt{3}}{2}i) (-\frac{1}{2} + \frac{\sqrt{3}}{2}i)$${z_2} = -\frac{1}{4} + \frac{\sqrt{3}}{4}i - \frac{\sqrt{3}}{4}i - \frac{3}{4} = -1$.

Let ${z_1}, {z_2}, {z_3}$ be three vertices of an equilateral triangle circumscribing the circle $|z| = \frac{1}{2}$. If ${z_1} = \frac{1}{2} + \frac{\sqrt{3}i}{2}$ and ${z_1}, {z_2}, {z_3}$ are in anticlockwise sense,then ${z_2}$ is

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