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

Given an equilateral triangle with vertices represented by complex numbers \( z_1, z_2, z_3 \) and its centroid \( z_0 \), we aim to compute \( \sum_{k=1}^{3} (z_k - z_0)^2 \).

1. The centroid \( z_0 \) of a triangle with vertices \( z_1, z_2, z_3 \) is defined as:

\[z_0 = \frac{z_1 + z_2 + z_3}{3}\]

1. This implies that the sum of the vertices is three times the centroid:

\[z_1 + z_2 + z_3 = 3z_0\]

1. The difference between each vertex and the centroid can be expressed as:

\[z_k - z_0 = z_k - \frac{z_1 + z_2 + z_3}{3}\]

1. For an equilateral triangle, the sum of the squared differences of its vertices from the centroid is:

\[\sum_{k=1}^{3} (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2\]

1. Due to the symmetric properties of an equilateral triangle with respect to its centroid, the terms \( (z_k - z_0) \) will cancel out when squared and summed. Specifically, the sum of these differences is zero:

\[(z_1 - z_0) + (z_2 - z_0) + (z_3 - z_0) = (z_1 + z_2 + z_3) - 3z_0 = 3z_0 - 3z_0 = 0\]

1. The sum of the squares of these deviations is therefore zero.

\[\sum_{k=1}^{3} (z_k - z_0)^2 = 0\]

The value is 0.