If z1, z2, z3 ∈ C are the vertices of an equilateral triangle, whose centroid is z0, then Σ(k=1 to 3

Q: Wait, why is the sum of squares of differences from the centroid zero? Does this mean the vertices are all at the origin?

Not at all! In the complex plane, the sum of vectors being zero simply means the centroid is the balance point. It is the same principle as the center of mass in physics being the point where the sum of moments is zero. The individual terms $(z_k - z_0)^2$ are complex numbers, and they can sum to zero without each term being zero individually.

Q: Can I just pick a specific triangle, like the one with vertices $1, \omega, \omega^2$, to solve this?

Absolutely! This is a brilliant strategy for competitive exams. If the result is independent of the specific triangle, it must hold for the simplest case. For $z_1=1, z_2=\omega, z_3=\omega^2$, the centroid $z_0$ is $0$. The sum becomes $1^2 + \omega^2 + \omega^4 = 1 + \omega^2 + \omega = 0$. It confirms our result perfectly.

Q: Is the identity $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$ always true for equilateral triangles?

Yes, it is a fundamental property. It arises because the roots of the equation $(z-z_1)(z-z_2)(z-z_3) = 0$ are the vertices, and for an equilateral triangle, the condition for the vertices to form such a shape is equivalent to this symmetric polynomial identity. It is a powerful tool in your JEE arsenal.

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

Options:

(A) 0

(B) 1

(C) $i$

(D) $-i$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are exploring the profound symmetry hidden within the complex plane.

Imagine you are standing on a vast, two-dimensional coordinate system. You have three points, $z_1, z_2, z_3$, forming a perfect, equilateral triangle. At the very heart of this triangle lies its centroid, $z_0$.

Our mission is to evaluate the sum of the squared differences from this centroid: $\sum_{k=1}^3 (z_k - z_0)^2$. It looks like a daunting algebraic task, but as we peel back the layers, you will see the elegance of the result.

### The Foundation: Defining the Centroid

First, let us ground ourselves. The centroid $z_0$ is the average of the vertices. Mathematically, we define it as:

$$z_0 = \frac{z_1 + z_2 + z_3}{3}$$

This gives us a crucial relationship: $z_1 + z_2 + z_3 = 3z_0$. Think of this as our anchor. Whenever we see the sum of the vertices, we can immediately replace it with $3z_0$.

### The Secret Weapon: The Equilateral Identity

Now, we need a tool to handle the squared terms. For any equilateral triangle in the complex plane, there is a beautiful, almost magical identity:

$$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$$

This identity is the key that unlocks the entire problem. It connects the sum of the squares of the vertices to the sum of their pairwise products. Keep this in your toolkit; it is a classic JEE favorite.

### The Expansion: The Algebraic Grind

Let us tackle the expression $\sum_{k=1}^3 (z_k - z_0)^2$. Expanding this, we get $(z_1 - z_0)^2 + (z_2 - z_0)^2 + (z_3 - z_0)^2$.

Using the expansion $(a-b)^2 = a^2 - 2ab + b^2$, we rewrite this as:

$$(z_1^2 - 2z_1z_0 + z_0^2) + (z_2^2 - 2z_2z_0 + z_0^2) + (z_3^2 - 2z_3z_0 + z_0^2)$$

Grouping the terms, we get $(z_1^2 + z_2^2 + z_3^2) - 2z_0(z_1 + z_2 + z_3) + 3z_0^2$. Now, substitute our anchor $z_1 + z_2 + z_3 = 3z_0$ into the middle term.

The expression becomes $(z_1^2 + z_2^2 + z_3^2) - 2z_0(3z_0) + 3z_0^2$, which simplifies to:

$$(z_1^2 + z_2^2 + z_3^2) - 6z_0^2 + 3z_0^2 = (z_1^2 + z_2^2 + z_3^2) - 3z_0^2$$

### The Climax: The Beautiful Cancellation

We have reduced the problem to $(z_1^2 + z_2^2 + z_3^2) - 3z_0^2$. Now, let us focus on $3z_0^2$. Since $z_0 = \frac{z_1 + z_2 + z_3}{3}$, we have:

$$3z_0^2 = 3 \left( \frac{z_1 + z_2 + z_3}{3} \right)^2 = \frac{(z_1 + z_2 + z_3)^2}{3}$$

Expanding the numerator, we get:

$$\frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1z_2 + z_2z_3 + z_3z_1)}{3}$$

Using our equilateral identity, we replace the pairwise product sum with the sum of squares:

$$\frac{z_1^2 + z_2^2 + z_3^2 + 2(z_1^2 + z_2^2 + z_3^2)}{3} = \frac{3(z_1^2 + z_2^2 + z_3^2)}{3} = z_1^2 + z_2^2 + z_3^2$$

Finally, substitute this back into our expression: $(z_1^2 + z_2^2 + z_3^2) - (z_1^2 + z_2^2 + z_3^2) = 0$. The terms cancel out perfectly, leaving us with the final result of **0**.

Animated Solution for Mathematics - Complex Numbers: If $z_{1}, z_{2}, z_{3} \in C$ 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

Select Answer:

Visualized Solution

The Equilateral Triangle

The Centroid

Centroid Formula

Equilateral Triangle Condition

The Target Expression

Expanding the Squares

Grouping the Terms

Substituting the Centroid

Simplifying the Expression

Expanding $3 z_{0}^{2}$

Applying the Equilateral Condition

Simplifying $3 z_{0}^{2}$

Final Result

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Foundation

The Secret Weapon

The Expansion

The Climax

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