If z_1, z_2, z_3 ∈ C are the vertices of an equilateral triangle, whose centroid is z_0 , then ∑_ k=1 ^3(z_ k -z_0)^2 is equal to

Correct answer: A. 0 aligned & z_1+z_2+z_3=3 z_0 & (z_1+z_2+z_3)^2=9 z_0^2 & z_1^2+z_2^2+z_3^2+2(z_1^2+z_2^2+z_3^2)=9 z_0^2 & z_1^2+z_2^2+z_3^2=3 z_6^2 aligned aligned & ∑_ k=1 ^3(z_ k -z_0)^2=(z_1-z_0)^2+(z_2-z_0)^2+(z_3-z_0)^2 & =z_1^2+z_2^2+z_3^2+3 z_0^2-2(z_1+z_2+z_3) z_0 & =6 z_0^2-6 z_0^2 & =0 aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Complex Number

If z_1, z_2, z_3 ∈ C are the vertices of an equilateral triangle,...

z_1, z_2, z_3 \in C

are the vertices of an equilateral triangle, whose centroid is

\mathrm{z}_0

, then $\sum_{\mathrm{k}=1}^3\left(\mathrm{z}_{\mathrm{k}}-\mathrm{z}_0\right)^2 is equal to

Options

A $0$
B $1$
Ci
D $-\mathrm{i}$

Correct answer

A. $0$

Step-by-step solution

\begin{aligned} & \mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3=3 \mathrm{z}_0 \\ & \left(\mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3\right)^2=9 \mathrm{z}_0^2 \\ & \Rightarrow \mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2+2\left(\mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2\right)=9 \mathrm{z}_0^2 \\ & \Rightarrow \mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2=3 \mathrm{z}_6^2\end{aligned}

\begin{aligned} & \sum_{\mathrm{k}=1}^3\left(\mathrm{z}_{\mathrm{k}}-\mathrm{z}_0\right)^2=\left(\mathrm{z}_1-\mathrm{z}_0\right)^2+\left(\mathrm{z}_2-\mathrm{z}_0\right)^2+\left(\mathrm{z}_3-\mathrm{z}_0\right)^2 \\ & \quad=\mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2+3 \mathrm{z}_0^2-2\left(\mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3\right) \mathrm{z}_0 \\ & =6 \mathrm{z}_0^2-6 \mathrm{z}_0^2 \\ & =0\end{aligned}

Practice Complex Number on Quantrex Academy →

More from Complex Number