JEE Main 2025 Apr 3 Shift 2, Mathematics Q15: Rotation And Geometric Applications

JEE Main2025Apr 3, Shift 2Mathematics

If $z_{1}, z_{2}, z_{3} \in C$ are the vertices of an equilateral triangle, whose centroid is $z_{0}$ , then $k = 1 \sum 3 (z_{k} - z_{0})^{2}$ is equal to

A
$0$
B
$π$
C
$i$
D
$- i$

Solution

By the centroid relation, $z_{1} + z_{2} + z_{3} = 3 z_{0}$ . Let $w_{k} = z_{k} - z_{0}$ . Then $w_{1} + w_{2} + w_{3} = 0$ .

For any three complex numbers with $w_{1} + w_{2} + w_{3} = 0$ ,

$(w_{1} + w_{2} + w_{3})^{2} = 0 \Rightarrow w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = - 2 (w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1})$ .

For an equilateral triangle centred at the origin, a standard identity is $w_{1} w_{2} + w_{2} w_{3} + w_{3} w_{1} = 0$ (the $w_{k}$ are equal-magnitude vectors at $12 0^{\circ}$ to each other, summing pairwise products to zero).

Hence $w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = 0$ .

$∴ k = 1 \sum 3 (z_{k} - z_{0})^{2} = 0$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q15: Rotation And Geometric Applications

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