[UPLIFTJEE]

If $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3} \in \mathrm{C}$ are the vertices of an equilateral triangle, whosecentroids $\mathrm{z}_{0}$, then $\sum_{\mathrm{k}=1}^{3}\left(\mathrm{z}_{\mathrm{k}}-\mathrm{z}_{0}\right)^{2}$ is equal to [JEE Main 03 April 2025 S II]
(A) 1
(B) i
(C) 0
(D) -i

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[UPLIFTJEE]

Solution

For equilateral triangle $\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}=\mathrm{z}_{1} \mathrm{z}_{2}+\mathrm{z}_{2} \mathrm{z}_{3}+\mathrm{z}_{3} \mathrm{z}_{1}$ Centroid $\left(\mathrm{z}_{0}\right) \equiv\left(\frac{\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}}{3}\right)$

\[ \begin{aligned} & \mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=3 \mathrm{z}_{0} \Rightarrow\left(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}\right)^{2}=9 \mathrm{z}_{0}^{2} \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}+2\left(\mathrm{z}_{1} \mathrm{z}_{2}+\mathrm{z}_{2} \mathrm{z}_{3}+\mathrm{z}_{3} \mathrm{z}_{1}\right)=9 \mathrm{z}_{0}^{2} \\ & \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}+2\left(\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}\right)=9 \mathrm{z}_{0}^{2} \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}=3 \mathrm{z}_{0}^{2} \\ & \sum_{\mathrm{k}=1}^{3}\left(\mathrm{z}_{\mathrm{k}}-\mathrm{z}_{0}\right)^{2}=\left(\mathrm{z}_{1}-\mathrm{z}_{0}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{0}\right)^{2}+\left(\mathrm{z}_{3}-\mathrm{z}_{0}\right)^{2}=\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}+\mathrm{z}_{3}^{2}+3 \mathrm{z}_{0}^{2}-2\left(\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}\right) \mathrm{z}_{0}=6 \mathrm{z}_{0}^{2}-6 \mathrm{z}_{0}^{2}=0 \end{aligned} \]

[UPLIFTJEE]

Answer

C

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