If $z_1, z_2, z_3 \in C$ are the vertices of an equilateral triangle, whose centroid is $\mathrm{z}_0$, then…

If $z_1, z_2, z_3 \in C$ 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
  1. $0$
  2. $1$
  3. i
  4. $-\mathrm{i}$

Solution

$\begin{aligned} & \mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3=3 \mathrm{z}_0 \\ & \left(\mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3\right)^2=9 \mathrm{z}_0^2 \\ & \Rightarrow \mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2+2\left(\mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2\right)=9 \mathrm{z}_0^2 \\ & \Rightarrow \mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2=3 \mathrm{z}_6^2\end{aligned}$
$\begin{aligned} & \sum_{\mathrm{k}=1}^3\left(\mathrm{z}_{\mathrm{k}}-\mathrm{z}_0\right)^2=\left(\mathrm{z}_1-\mathrm{z}_0\right)^2+\left(\mathrm{z}_2-\mathrm{z}_0\right)^2+\left(\mathrm{z}_3-\mathrm{z}_0\right)^2 \\ & \quad=\mathrm{z}_1^2+\mathrm{z}_2^2+\mathrm{z}_3^2+3 \mathrm{z}_0^2-2\left(\mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3\right) \mathrm{z}_0 \\ & =6 \mathrm{z}_0^2-6 \mathrm{z}_0^2 \\ & =0\end{aligned}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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