SATHEE: Complex Numbers 4 Question 18

Complex Numbers 4 Question 18

18.

Let the complex numbers $z_{1}, z_{2}$ and $z_{3}$ be the vertices of an equilateral triangle. Let $z_{0}$ be the circumcentre of the triangle. Then, prove that $z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3 z_{0}^{2}$ .

(1981, 4M)

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Solution:

Since, $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle.

$∴$ Circumcentre $(z_{0}) =$ Centroid $\frac{z_{1} + z_{2} + z_{3}}{3}$

Also, for equilateral triangle

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

On squaring Eq. (i), we get

$\begin{aligned} 9 z_{0}^{2} & = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}) \\ \Rightarrow 9 z_{0}^{2} & = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) [from Eq. (ii)] \\ \Rightarrow 3 z_{0}^{2} & = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} \end{aligned}$

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Complex Numbers 4 Question 18

18.

Solution:

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