JEE Mains · Maths · STD 11 - 4.1 complex nubers
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
- A \(0\)
- B \(1\)
- C i
- D \(-\mathrm{i}\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
\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…
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