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

To solve the problem, we have an equilateral triangle with vertices represented as complex numbers \( z_1, z_2, z_3 \), and the centroid of this triangle is \( z_0 \). We need to find the value of \( \sum_{k=1}^{3} (z_k - z_0)^2 \).

1. First, recall that the centroid of a triangle with vertices \( z_1, z_2, z_3 \) in the complex plane is given by:

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

1. Each vertex can be expressed in terms of the centroid as:

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

1. We need to compute the sum of the square of these differences:

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

1. Using the identity of an equilateral triangle, we know:

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

1. Substituting this into our equation, we see:

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

1. Therefore, each term evaluates to zero since the sum of deviations of the vertices of an equilateral triangle from its centroid squared results in zero:

\[(z_k - z_0) \text{ symmetrically cancels out across } z_1, z_2, \text{ and } z_3\]

1. This property arises due to the symmetry of an equilateral triangle.
2. The sum \(\sum_{k=1}^{3} (z_k - z_0)^2\) evaluates to:

\[0\]

Thus, the correct answer is: 0

Answer 2

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: \[ \sum_{k=1}^{3} (z_k - z_0)^2 = z_1^2 + z_2^2 + z_3^2 - 3z_0^2 = 3z_0^2 - 3z_0^2 = 0 \]