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$$ \begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned} $$

Let $z_0$ is origin $\Rightarrow z_1, z_2, z_3$ lies on a circle having $\left|z_0-z_i\right|=R$

$$ \begin{aligned} & \therefore z_1=R e^{i 2 \pi / 3} z_2=R e^{i 4 \pi / 3} z_3=R e^{i 6 \pi / 3} \\ & \Rightarrow z_1^2+z_2^2+z_3^2=R^2\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right] \\ & =0 \\ & \therefore \sum_{k=1}^3\left(z_k-z_0\right)^2=0 \end{aligned} $$

19:T493,

JEE Main 2025 (Online) 3rd April Evening Shift Mathematics - Complex Numbers Question 20 Hindi Explanation

$$ \begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned} $$

माना $z_0$ मूल बिंदु है $\Rightarrow z_1, z_2, z_3$ एक वृत्त पर स्थित हैं जिसकी त्रिज्या $\left|z_0-z_i\right|=R$ है

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