Logo
Jeeify
MCQMediumJEE 2025Argand Plane & Geometry

JEE Mathematics 2025 Question with Solution

If z1,z2,z3∈Cz_1, z_2, z_3 \in \mathbb{C} are the vertices of an equilateral triangle, whose centroid is z0z_0, then ∑k=13(zk−z0)2\sum_{k=1}^{3} (z_k - z_0)^2 is equal to

  • A

    00

  • B

    22

  • C

    3i3i

  • D

    −i-i

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: z1,z2,z3z_1, z_2, z_3 are the vertices of an equilateral triangle and the centroid is z0z_0.

Find: ∑k=13(zk−z0)2\sum_{k=1}^{3} (z_k-z_0)^2.

Using the centroid formula,

z0=z1+z2+z33z_0 = \frac{z_1+z_2+z_3}{3}

so

z1+z2+z3=3z0z_1+z_2+z_3 = 3z_0

Now expand the required sum:

(z1−z0)2+(z2−z0)2+(z3−z0)2(z_1-z_0)^2+(z_2-z_0)^2+(z_3-z_0)^2 =(z12+z22+z32)−2z0(z1+z2+z3)+3z02= (z_1^2+z_2^2+z_3^2)-2z_0(z_1+z_2+z_3)+3z_0^2

Substituting z1+z2+z3=3z0z_1+z_2+z_3=3z_0,

∑k=13(zk−z0)2=z12+z22+z32−6z02+3z02\sum_{k=1}^{3} (z_k-z_0)^2 = z_1^2+z_2^2+z_3^2-6z_0^2+3z_0^2 =z12+z22+z32−3z02= z_1^2+z_2^2+z_3^2-3z_0^2

For the vertices of an equilateral triangle, the relation

z12+z22+z32=z1z2+z2z3+z3z1z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1

holds. Also,

(z1+z2+z3)2=z12+z22+z32+2(z1z2+z2z3+z3z1)(z_1+z_2+z_3)^2 = z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)

Since z1+z2+z3=3z0z_1+z_2+z_3=3z_0,

9z02=z12+z22+z32+2(z1z2+z2z3+z3z1)9z_0^2 = z_1^2+z_2^2+z_3^2+2(z_1z_2+z_2z_3+z_3z_1)

Using z12+z22+z32=z1z2+z2z3+z3z1z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1,

9z02=3(z12+z22+z32)9z_0^2 = 3(z_1^2+z_2^2+z_3^2)

Therefore,

z12+z22+z32=3z02z_1^2+z_2^2+z_3^2 = 3z_0^2

Substitute back:

∑k=13(zk−z0)2=3z02−3z02=0\sum_{k=1}^{3} (z_k-z_0)^2 = 3z_0^2-3z_0^2 = 0

Therefore, the correct option is A.

Symmetry-Based Shortcut

Given: z1,z2,z3z_1, z_2, z_3 form an equilateral triangle with centroid z0z_0.

Find: ∑k=13(zk−z0)2\sum_{k=1}^{3} (z_k-z_0)^2.

Shift the origin to the centroid by taking

wk=zk−z0w_k = z_k-z_0

Then

w1+w2+w3=0w_1+w_2+w_3=0

and the three points w1,w2,w3w_1,w_2,w_3 are equally spaced by angle 2π3\frac{2\pi}{3} about the origin.

Hence they can be written as

w,  wω,  wω2w,\; w\omega,\; w\omega^2

where ω3=1\omega^3=1 and 1+ω+ω2=01+\omega+\omega^2=0.

Therefore,

w12+w22+w32=w2(1+ω2+ω4)w_1^2+w_2^2+w_3^2 = w^2\left(1+\omega^2+\omega^4\right)

Since ω4=ω\omega^4=\omega,

1+ω+ω2=01+\omega+\omega^2=0

so the sum is 00.

Thus,

∑k=13(zk−z0)2=0\sum_{k=1}^{3} (z_k-z_0)^2 = 0

Therefore, the correct option is A.

Common mistakes

  • Using only the centroid relation z1+z2+z3=3z0z_1+z_2+z_3=3z_0 and concluding the squared sum is automatically zero. The sum of deviations is zero, but the sum of their squares is a different expression. Expand the squares or use the equilateral-triangle symmetry property.

  • Expanding (zk−z0)2(z_k-z_0)^2 incorrectly by missing the middle term or handling the three z02z_0^2 terms wrongly. Write the full expansion carefully and then combine all three terms systematically.

  • Forgetting the special identity for an equilateral triangle, namely z12+z22+z32=z1z2+z2z3+z3z1z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1. Without this extra symmetry relation, the expression cannot be reduced to zero.

Practice more Argand Plane & Geometry questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Argand Plane & Geometry

Related questions