The centre of a square ABCD is at z=0, A is `z_(1)`. Then, the centroid of `/_\ABC` is (where, `i=sqrt(-1)`)

Options

A. `z_(1)(cos pi +- isinpi)`
B. `z_(1)/3(cos pi +- isinpi)`
C. `z_(1)(cos (pi/(2)) +- isin(pi/2))`
D. `z_(1)/3(cos (pi/(2)) +- isin(pi/2))`

Answer

To find the centroid of triangle ABC where the center of square ABCD is at \( z = 0 \) and point A is at \( z_1 \), we can follow these steps:

### Step 1: Identify the points of the square
Let the vertices of square ABCD be represented in the complex plane. Since the center is at \( z = 0 \), we can denote the points as follows:
- \( A = z_1 \)
- \( B = z_2 \)
- \( C = z_3 \)
- \( D = z_4 \)

### Step 2: Express points B and C in terms of A
Since the square is symmetric and centered at the origin, we can find the other vertices using rotation:
- The point \( B \) can be obtained by rotating \( A \) by \( 90^\circ \) counterclockwise:
\[ z_2 = z_1 e^{i \frac{\pi}{2}} = z_1 i \]
- The point \( C \) can be obtained by rotating \( A \) by \( 180^\circ \):
\[ z_3 = z_1 e^{i \pi} = -z_1 \]

### Step 3: Calculate the centroid of triangle ABC
The centroid \( G \) of triangle ABC is given by the formula:
\[ G = \frac{z_1 + z_2 + z_3}{3} \]
Substituting the values of \( z_2 \) and \( z_3 \):
\[ G = \frac{z_1 + z_1 i - z_1}{3} \]
This simplifies to:
\[ G = \frac{z_1 i}{3} \]

### Step 4: Final expression for the centroid
Thus, the centroid of triangle ABC is:
\[ G = \frac{z_1}{3} i \]