Question

If \( z_1, z_2, z_3 \in \mathbb{C} \) are the vertices of an equilateral triangle, whose centroid is \( z_0 \), then \( \sum_{k=1}^{3} (z_k - z_0)^2 \) is equal to

• A. 0
• B. 2
• C. \( 3i \)
• D. \( -i \)

Hint:

For the vertices \( z_1, z_2, z_3 \) of an equilateral triangle with centroid \( z_0 \), the relation \( z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \) holds. Also, the centroid is the average of the vertices: \( z_0 = \frac{z_1 + z_2 + z_3}{3} \). Use these properties to simplify the expression \( \sum_{k=1}^{3} (z_k - z_0)^2 \).

Answer 1

The Correct Option is A

The centroid \( z_0 \) of a triangle with vertices \( z_1, z_2, z_3 \) is given by: \[ z_0 = \frac{z_1 + z_2 + z_3}{3} \] From this, we have: \[ z_1 + z_2 + z_3 = 3z_0 \] We need to find the value of \( \sum_{k=1}^{3} (z_k - z_0)^2 \), which is: \[ (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2 \] Expanding the terms: \[ (z_1^2 - 2z_1 z_0 + z_0^2) + (z_2^2 - 2z_2 z_0 + z_0^2) + (z_3^2 - 2z_3 z_0 + z_0^2) \] \[ = (z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2 \] Substitute \( z_1 + z_2 + z_3 = 3z_0 \) into the expression: \[ = (z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2 \] \[ = z_1^2 + z_2^2 + z_3^2 - 6z_0^2 + 3z_0^2 \] \[ = z_1^2 + z_2^2 + z_3^2 - 3z_0^2 \] Now, consider \( (z_1 + z_2 + z_3)^2 \): \[ (z_1 + z_2 + z_3)^2 = z_1^2 + z_2^2 + z_3^2 + 2(z_1 z_2 + z_2 z_3 + z_3 z_1) \] Also, \( (z_1 + z_2 + z_3)^2 = (3z_0)^2 = 9z_0^2 \). So, \( z_1^2 + z_2^2 + z_3^2 + 2(z_1 z_2 + z_2 z_3 + z_3 z_1) = 9z_0^2 \). For an equilateral triangle with centroid \( z_0 \), we also have the property: \[ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \] Substituting this into the previous equation: \[ (z_1^2 + z_2^2 + z_3^2) + 2(z_1^2 + z_2^2 + z_3^2) = 9z_0^2 \] \[ 3(z_1^2 + z_2^2 + z_3^2) = 9z_0^2 \] \[ z_1^2 + z_2^2 + z_3^2 = 3z_0^2 \] Now substitute this back into the expression for the sum: \[ \sum_{k=1}^{3} (z_k - z_0)^2 = z_1^2 + z_2^2 + z_3^2 - 3z_0^2 = 3z_0^2 - 3z_0^2 = 0 \]