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. (a)`z_(1)(cos pi +- isinpi)`
B. (b)`z_(1)/3(cos pi +- isinpi)`
C. (c)`z_(1)(cos (pi/(2)) +- isin(pi/2))`
D. (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 represented by \( z_1 \), we can follow these steps:

### Step 1: Define the vertices of the square
Let the vertices of the square ABCD be represented in the complex plane. Since the center is at \( z = 0 \), we can denote:
- \( A = z_1 \)
- \( B = z_1 e^{i\frac{\pi}{2}} \) (which corresponds to a 90-degree rotation of A)
- \( C = z_1 e^{i\pi} \) (which corresponds to a 180-degree rotation of A)
- \( D = z_1 e^{i\frac{3\pi}{2}} \) (which corresponds to a 270-degree rotation of A)

### Step 2: Express the points in terms of \( z_1 \)
Using Euler's formula, we can express the points as:
- \( A = z_1 \)
- \( B = z_1 \cdot i \) (since \( e^{i\frac{\pi}{2}} = i \))
- \( C = -z_1 \) (since \( e^{i\pi} = -1 \))
- \( D = -z_1 \cdot i \) (since \( e^{i\frac{3\pi}{2}} = -i \))

### Step 3: Calculate the centroid of triangle ABC
The centroid \( G \) of triangle ABC is given by the formula:
\[ G = \frac{A + B + C}{3} \]
Substituting the values of A, B, and C:
\[ G = \frac{z_1 + (z_1 \cdot i) + (-z_1)}{3} \]

### Step 4: Simplify the expression
Now, simplifying the expression:
\[ G = \frac{z_1 + z_1 \cdot i - z_1}{3} \]
This simplifies to:
\[ G = \frac{z_1 \cdot i}{3} \]

### Final Result
Thus, the centroid of triangle ABC is:
\[ G = \frac{z_1 \cdot i}{3} \]