Question

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 \[{z_1}^2 + {z_2}^2 + {z_3}^2 = \] [IIT \[1981\]]
A) \[{z_0}^2\]
B) \[ - {z_0}^2\]
C) \[3{z_0}^2\]
D) \[ - 3{z_0}^2\]

Answer 1

Hint: in this question we have to value of\[{z_1}^2 + {z_2}^2 + {z_3}^2\]. Use the formula of centroid of equilateral triangle. Circumcentre is same as centroid. Use property of equilateral triangle in complex number.

Formula Used: Centroid of triangle is given by
\[{z_0} = \dfrac{{{z_1} + {z_2} + {z_3}}}{3}\]
property of equilateral triangle
\[{z_1}^2 + {z_2}^2 + {z_3}^2 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}\]
Where
\[{z_1}\],\[{z_2}\] and \[{z_3}\]are the vertices of triangle

Complete step by step solution: Given:\[{z_1}\],\[{z_2}\] and \[{z_3}\]are the vertices of equilateral triangle
We know that
Centroid of triangle is given by
\[{z_0} = \dfrac{{{z_1} + {z_2} + {z_3}}}{3}\]……………………………. (i)
Where
\[{z_1}\],\[{z_2}\] and \[{z_3}\]are the vertices of triangle
Squaring both side of equation (i)
\[9{z_0}^2 = {z_1}^2 + {z_2}^2 + {z_3}^2 + 2({z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1})\]……………………………… (ii)
Now from property of equilateral triangle
\[{z_1}^2 + {z_2}^2 + {z_3}^2 = {z_1}{z_2} + {z_2}{z_3} + {z_3}{z_1}\]………………………………………….. (iii)
From equation (ii) and (iii)
\[9{z_0}^2 = {z_1}^2 + {z_2}^2 + {z_3}^2 + 2({z_1}^2 + {z_2}^2 + {z_3}^2)\]
\[9{z_0}^2 = 3({z_1}^2 + {z_2}^2 + {z_3}^2)\]
\[{z_1}^2 + {z_2}^2 + {z_3}^2 = 3{z_0}^2\]

Option ‘C’ is correct

Note: Here we must remember circumcentre and centroid are same for equilateral triangle and property of equilateral triangle. Complex number is a number which is a union of real and imaginary number.