`z_(1),z_(2),z_(3)` are the vertices of an equilateral triangle taken in counter clockwise direction. If its circumcenter is at `(1-2i)` and `(z_(1)=2+i)`, then `z_(2)=`

Options

A. `(1-3sqrt(3))/2+(sqrt(3)-7)/(2)i`
B. `(1+3sqrt(3))/(2) -(7+sqrt(3))/(2)j`
C. `(1+3sqrt(3))/(2), (sqrt(3)-7)/(2)i`
D. `(1+3sqrt(3))/(2) +(7+sqrt(3))/(2)`i

Explanation

To find the vertex \( z_2 \) of the equilateral triangle given the circumcenter and one vertex, we can follow these steps:

### Step 1: Identify the given information
We know:
- The circumcenter \( O = 1 - 2i \)
- The vertex \( z_1 = 2 + i \)

### Step 2: Calculate the vector from the circumcenter to \( z_1 \)
The vector \( OA \) from the circumcenter \( O \) to vertex \( z_1 \) can be calculated as:
\[ OA = z_1 - O = (2 + i) - (1 - 2i) = 2 + i - 1 + 2i = 1 + 3i \]

### Step 3: Rotate the vector \( OA \) by \( \frac{2\pi}{3} \)
To find the next vertex \( z_2 \), we need to rotate the vector \( OA \) by \( \frac{2\pi}{3} \) radians (which is \( 120^\circ \)) counterclockwise. The rotation can be done using the multiplication by the complex number \( e^{i\frac{2\pi}{3}} \):
\[ e^{i\frac{2\pi}{3}} = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]
Thus, we calculate \( OB \):
\[ OB = OA \cdot e^{i\frac{2\pi}{3}} = (1 + 3i) \cdot \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \]

### Step 4: Perform the multiplication
Calculating \( OB \):
\[ OB = 1 \cdot \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) + 3i \cdot \left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \]
\[ = -\frac{1}{2} + i\frac{\sqrt{3}}{2} - \frac{3}{2}i - 3\frac{\sqrt{3}}{2} \]
\[ = -\frac{1}{2} - \frac{3\sqrt{3}}{2} + i\left(\frac{\sqrt{3}}{2} - \frac{3}{2}\right) \]

### Step 5: Calculate \( z_2 \)
Now, we add the vector \( OB \) to the circumcenter \( O \):
\[ z_2 = O + OB = (1 - 2i) + \left(-\frac{1}{2} - \frac{3\sqrt{3}}{2} + i\left(\frac{\sqrt{3}}{2} - \frac{3}{2}\right)\right) \]
\[ = \left(1 - \frac{1}{2} - \frac{3\sqrt{3}}{2}\right) + i\left(-2 + \frac{\sqrt{3}}{2} - \frac{3}{2}\right) \]
\[ = \frac{1}{2} - \frac{3\sqrt{3}}{2} + i\left(-\frac{7}{2} + \frac{\sqrt{3}}{2}\right) \]

### Final Answer
Thus, the coordinates of \( z_2 \) are:
\[ z_2 = \frac{1 - 3\sqrt{3}}{2} + i\left(\frac{\sqrt{3} - 7}{2}\right) \]

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