To solve this problem, we need to determine the next vertex \( z_2 \) of the equilateral triangle given that \( z_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2} \) is one of the vertices and the vertices are described in an anticlockwise direction.
An equilateral triangle inscribed in a circle (i.e., circumcircle) can also be considered to be rotated by an angle of \( \frac{2\pi}{3} \) radians (120 degrees) about the center of the circle to reach from one vertex to the next.
The given circle is described by the equation \( |z| = \frac{1}{2} \). This implies that the radius of the circumcircle of the triangle is \( \frac{1}{2} \).
Since \( z_1 = \frac{1}{2} + i\frac{\sqrt{3}}{2} \), we need to rotate this point by \( \frac{2\pi}{3} \) anticlockwise about the circle's center to find \( z_2 \).
The rotation transformation in the complex plane can be represented by multiplying with \( e^{i\frac{2\pi}{3}} \). Therefore, we calculate:
\(z_2 = z_1 \cdot e^{i\frac{2\pi}{3}}\)
The complex exponential \( e^{i\frac{2\pi}{3}} \) can be expressed using Euler's formula as:
\(e^{i\frac{2\pi}{3}} = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right)\)
\(\Rightarrow e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
Now let's do the multiplication:
\(z_2 = \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \cdot \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)\)
Performing the multiplication, we use the distributive property:
\(z_2 = \left(\frac{1}{2} \cdot -\frac{1}{2}\right) + \left(\frac{1}{2} \cdot i\frac{\sqrt{3}}{2}\right) + \left(i\frac{\sqrt{3}}{2} \cdot -\frac{1}{2}\right) + \left(i\frac{\sqrt{3}}{2} \cdot i\frac{\sqrt{3}}{2}\right)\)
This gives:
\(z_2 = -\frac{1}{4} + i\frac{\sqrt{3}}{4} - i\frac{\sqrt{3}}{4} - \frac{3}{4}\)
Simplifying, the imaginary terms \( i\frac{\sqrt{3}}{4} - i\frac{\sqrt{3}}{4} \) cancel each other:
\(z_2 = -\frac{1}{4} - \frac{3}{4} = -1\)
Therefore, the second vertex of the triangle, \( z_2 \), is \( -1 \).
Thus, the correct answer is:
-1