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.
$\therefore$ Circumcentre $\left(z _0\right)=$ 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\left(z _1 z _2+z _2 z _3+z _3 z _1\right) \\ \Rightarrow 9 z _0^{2} & =z _1^{2}+z _2^{2}+z _3^{2}+2\left(z _1^{2}+z _2^{2}+z _3^{2}\right) \quad \text { [from Eq. (ii)] } \\ \Rightarrow 3 z _0^{2} & =z _1^{2}+z _2^{2}+z _3^{2} \end{aligned} $
