Key Concepts and Formulas

• Centroid of a Triangle: The centroid of a triangle with vertices $z_1, z_2, z_3$ is given by $z_0 = \frac{z_1 + z_2 + z_3}{3}$.
• Equilateral Triangle Property: If $z_1, z_2, z_3$ form an equilateral triangle, and $\omega = e^{i2\pi/3}$ is a cube root of unity, then $z_1 + z_2\omega + z_3\omega^2 = 0$ or $z_1 + z_2\omega^2 + z_3\omega = 0$ after suitable rotation.
• Cube Roots of Unity: $1 + \omega + \omega^2 = 0$, where $\omega = e^{i2\pi/3}$.

Step-by-Step Solution

Step 1: Simplify the Expression Using Centroid Property

• What we are doing: We aim to simplify the expression $\sum_{k=1}^3 (z_k - z_0)^2$ using the fact that $z_0$ is the centroid.
• Why we are doing it: Substituting the expression for $z_0$ will allow us to rewrite the sum in terms of only $z_1, z_2,$ and $z_3$, which might reveal hidden relationships.
• We have $z_0 = \frac{z_1 + z_2 + z_3}{3}$. Thus, $z_k - z_0 = z_k - \frac{z_1 + z_2 + z_3}{3}$ for $k = 1, 2, 3$.
• The sum becomes: $\sum_{k=1}^3 (z_k - z_0)^2 = (z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$ $= \left(z_1 - \frac{z_1 + z_2 + z_3}{3}\right)^2 + \left(z_2 - \frac{z_1 + z_2 + z_3}{3}\right)^2 + \left(z_3 - \frac{z_1 + z_2 + z_3}{3}\right)^2$ $= \left(\frac{2z_1 - z_2 - z_3}{3}\right)^2 + \left(\frac{2z_2 - z_1 - z_3}{3}\right)^2 + \left(\frac{2z_3 - z_1 - z_2}{3}\right)^2$ $= \frac{1}{9} \left[ (2z_1 - z_2 - z_3)^2 + (2z_2 - z_1 - z_3)^2 + (2z_3 - z_1 - z_2)^2 \right]$ $= \frac{1}{9} \left[ 4z_1^2 + z_2^2 + z_3^2 - 4z_1z_2 - 4z_1z_3 + 2z_2z_3 + 4z_2^2 + z_1^2 + z_3^2 - 4z_2z_1 - 4z_2z_3 + 2z_1z_3 + 4z_3^2 + z_1^2 + z_2^2 - 4z_3z_1 - 4z_3z_2 + 2z_1z_2 \right]$ $= \frac{1}{9} \left[ 6(z_1^2 + z_2^2 + z_3^2) - 6(z_1z_2 + z_1z_3 + z_2z_3) \right]$ $= \frac{2}{3} \left[ z_1^2 + z_2^2 + z_3^2 - z_1z_2 - z_1z_3 - z_2z_3 \right]$

Step 2: Use Equilateral Triangle Property

• What we are doing: We leverage the equilateral triangle property to simplify the expression further.
• Why we are doing it: The equilateral triangle property provides a relationship between $z_1, z_2, z_3$ that we can substitute into our simplified expression.
• Since $z_1, z_2, z_3$ form an equilateral triangle, we can say $z_1 + \omega z_2 + \omega^2 z_3 = 0$ (or some cyclic permutation of this).
• Consider $(z_1 + \omega z_2 + \omega^2 z_3)^2 = 0$. $z_1^2 + \omega^2 z_2^2 + \omega^4 z_3^2 + 2\omega z_1z_2 + 2\omega^2 z_1z_3 + 2\omega^3 z_2z_3 = 0$ Since $\omega^3 = 1$ and $\omega^4 = \omega$, $z_1^2 + \omega^2 z_2^2 + \omega z_3^2 + 2\omega z_1z_2 + 2\omega^2 z_1z_3 + 2 z_2z_3 = 0$
• Similarly, consider $(z_1 + \omega^2 z_2 + \omega z_3)^2 = 0$. $z_1^2 + \omega^4 z_2^2 + \omega^2 z_3^2 + 2\omega^2 z_1z_2 + 2\omega z_1z_3 + 2\omega^3 z_2z_3 = 0$ $z_1^2 + \omega z_2^2 + \omega^2 z_3^2 + 2\omega^2 z_1z_2 + 2\omega z_1z_3 + 2 z_2z_3 = 0$
• Adding the two squared equations: $2z_1^2 + (\omega^2 + \omega)z_2^2 + (\omega + \omega^2)z_3^2 + 2(\omega + \omega^2)z_1z_2 + 2(\omega^2 + \omega)z_1z_3 + 4z_2z_3 = 0$ $2z_1^2 - z_2^2 - z_3^2 - 2z_1z_2 - 2z_1z_3 + 4z_2z_3 = 0$ This does not seem to lead to a direct solution.

Step 3: Alternative approach using $z_1 + z_2 + z_3 = 3z_0$

• What we are doing: Try to express everything in terms of $z_0$ and use the equilateral triangle property
• Why we are doing it: This can help to simplify the expression and arrive at a conclusive answer.

Since $z_1, z_2, z_3$ are vertices of an equilateral triangle, $(z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0$. Expanding this, we get $2(z_1^2 + z_2^2 + z_3^2) - 2(z_1z_2 + z_2z_3 + z_3z_1) = 0$. Thus, $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$.

Substituting this into the result from Step 1: $\frac{2}{3} \left[ z_1^2 + z_2^2 + z_3^2 - (z_1z_2 + z_1z_3 + z_2z_3) \right] = \frac{2}{3} [0] = 0$

Step 4: Conclusion

• What we are doing: State the final answer.
• Why we are doing it: To provide a clear, definitive answer to the original question.
• Thus, $\sum_{k=1}^3 (z_k - z_0)^2 = 0$.

Common Mistakes & Tips

• Remember the correct formula for the centroid.
• Utilize the properties of equilateral triangles in the complex plane, particularly the relationship between the vertices and the cube roots of unity.
• Be careful with algebraic manipulations and expansions.

Summary

By using the definition of the centroid and the properties of equilateral triangles, we simplified the given expression and found that the sum of the squares of the differences between the vertices and the centroid is equal to zero.

The final answer is $\boxed{0}$, which corresponds to option (A).