Question 22 of 455 Medium MCQ
\[If\,\,{z_1},{z_2},{z_3} \in \,\,are\,\,the\,\,vertices\,\,of\,\,an\,\,equilateral\,\,triangle,\,\,whose\,\,centroid\,\,is\,\,{z_0},\,\,then\,\,\sum\limits_{k = 1}^3 {{{\left( {{z_k} - {z_0}} \right)}^2}\,is\,\,equal\,\,to} \]
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\[ z 0 = z 1 + z 2 + z 3 ∑ k = 1 3 ( z k − z 0 ) 2 = ( z 1 − z 0 ) 2 + ( z 2 − z 0 ) 2 + ( z 3 − z 0 ) 2 \begin{aligned} & z_0=z_1+z_2+z_3 \\ & \sum_{k=1}^3\left(z_k-z_0\right)^2=\left(z_1-z_0\right)^2+\left(z_2-z_0\right)^2+\left(z_3-z_0\right)^2 \end{aligned} z 0 = z 1 + z 2 + z 3 k = 1 ∑ 3 ( z k − z 0 ) 2 = ( z 1 − z 0 ) 2 + ( z 2 − z 0 ) 2 + ( z 3 − z 0 ) 2 \]
Let z 0 z_0 z 0 is origin ⇒ z 1 , z 2 , z 3 \Rightarrow z_1, z_2, z_3 ⇒ z 1 , z 2 , z 3 lies on a circle having ∣ z 0 − z i ∣ = R \left|z_0-z_i\right|=R ∣ z 0 − z i ∣ = R
\[ ∴ z 1 = R e i 2 π / 3 z 2 = R e i 4 π / 3 z 3 = R e i 6 π / 3 ⇒ z 1 2 + z 2 2 + z 3 2 = R 2 [ e i 4 π / 3 + e i 8 π / 3 + e i 12 π / 3 ] = 0 ∴ ∑ k = 1 3 ( z k − z 0 ) 2 = 0 \begin{aligned} & \therefore z_1=R e^{i 2 \pi / 3} z_2=R e^{i 4 \pi / 3} z_3=R e^{i 6 \pi / 3} \\ & \Rightarrow z_1^2+z_2^2+z_3^2=R^2\left[e^{i 4 \pi / 3}+e^{i 8 \pi / 3}+e^{i 12 \pi / 3}\right] \\ & =0 \\ & \therefore \sum_{k=1}^3\left(z_k-z_0\right)^2=0 \end{aligned} ∴ z 1 = R e i 2 π /3 z 2 = R e i 4 π /3 z 3 = R e i 6 π /3 ⇒ z 1 2 + z 2 2 + z 3 2 = R 2 [ e i 4 π /3 + e i 8 π /3 + e i 12 π /3 ] = 0 ∴ k = 1 ∑ 3 ( z k − z 0 ) 2 = 0 \] Answer the question to unlock