If $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3} \in \mathrm{C}$ 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

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Complex Numbers

z_{1}, z_{2}, z_{3} \in C

are the vertices of an equilateral triangle, whose centroid is

z_{0}

, then

\sum_{k = 1}^{3} (z_{k} - z_{0})^{2}

is equal to

Answer: (A)

The centroid gives

z_{1} + z_{2} + z_{3} = 3 z_{0}

. Expanding,

\sum_{k = 1}^{3} (z_{k} - z_{0})^{2} = \sum z_{k}^{2} - 2 z_{0} \sum z_{k} + 3 z_{0}^{2} = \sum z_{k}^{2} - 6 z_{0}^{2} + 3 z_{0}^{2} = \sum z_{k}^{2} - 3 z_{0}^{2}

. For an equilateral triangle

z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}

; combining with

(z_{1} + z_{2} + z_{3})^{2} = 9 z_{0}^{2}

gives

\sum z_{k}^{2} + 2 \sum z_{i} z_{j} = 9 z_{0}^{2}

, hence

3 \sum z_{k}^{2} = 9 z_{0}^{2}

and

\sum z_{k}^{2} = 3 z_{0}^{2}

. Substituting,

\sum_{k = 1}^{3} (z_{k} - z_{0})^{2} = 3 z_{0}^{2} - 3 z_{0}^{2}

, which equals

0

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option