lf `z_1,z_2,z_3` are vertices of an equilateral triangle inscribed in the circle `|z| = 2` and if `z_1 = 1 + iotasqrt3` , then

Options

A. `z_(2)=-2,z_(3)=1-isqrt(3)`
B. `z_(2)=2,z_(3)=1-isqrt(3)`
C. `z_(2)=-2,z_(3)=-1-isqrt(3)`
D. `z_(2)=1-isqrt(3),z_(3)=1-isqrt(3)`

Answer

To solve the problem, we need to find the other two vertices \( z_2 \) and \( z_3 \) of the equilateral triangle inscribed in the circle \( |z| = 2 \) given that \( z_1 = 1 + i\sqrt{3} \).

### Step-by-Step Solution:

1. **Identify the Radius and Center of the Circle:**
The circle is given by \( |z| = 2 \), which means the radius \( r = 2 \) and the center is at the origin \( (0, 0) \).

2. **Convert \( z_1 \) to Polar Form:**
We can express \( z_1 \) in polar form. The modulus of \( z_1 \) is:
\[ |z_1| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]
The argument \( \theta_1 \) of \( z_1 \) is:
\[ \theta_1 = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \]
Thus, in polar form:
\[ z_1 = 2 \left( \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} \right) \]

3. **Find \( z_2 \) and \( z_3 \) Using Rotations:**
Since \( z_2 \) and \( z_3 \) are vertices of the equilateral triangle, they can be found by rotating \( z_1 \) by \( 120^\circ \) and \( 240^\circ \) respectively.

- **Finding \( z_2 \):**
\[ \theta_2 = \theta_1 + \frac{2\pi}{3} = \frac{\pi}{3} + \frac{2\pi}{3} = \pi \]
Thus,
\[ z_2 = 2 \left( \cos \pi + i \sin \pi \right) = 2(-1 + 0i) = -2 \]

- **Finding \( z_3 \):**
\[ \theta_3 = \theta_1 + \frac{4\pi}{3} = \frac{\pi}{3} + \frac{4\pi}{3} = \frac{5\pi}{3} \]
Thus,
\[ z_3 = 2 \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right) \]
We know that:
\[ \cos \frac{5\pi}{3} = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos \frac{\pi}{3} = \frac{1}{2} \]
\[ \sin \frac{5\pi}{3} = \sin\left(2\pi - \frac{\pi}{3}\right) = -\sin \frac{\pi}{3} = -\frac{\sqrt{3}}{2} \]
Therefore,
\[ z_3 = 2 \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = 1 - i\sqrt{3} \]

4. **Final Results:**
The vertices of the equilateral triangle are:
\[ z_1 = 1 + i\sqrt{3}, \quad z_2 = -2, \quad z_3 = 1 - i\sqrt{3} \]

### Summary:
Thus, we find:
- \( z_2 = -2 \)
- \( z_3 = 1 - i\sqrt{3} \)