Let the complex number $z_{1}, z_{2}, z_{3}$ be the varticles of an equilateral triangle . let $z_{0}$ be the cirumcental of the triangle . prove that ${z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 z_{0}^{2}$

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Solution

let the vertices of the $t r i a n g l e A B C$ be represented by $z_{1}, z_{2}, z_{3}$ then by the rotation in anti clock about A , B , we get
$A C = A B e^{π i / 3},, B A = A C e^{π i / 3},$
$(z_{3} - z_{1}) = (z_{2} - z_{1}) e^{π i / 3} (z_{1} - z_{2}) = (z_{3} - z_{2}) e^{π i / 3}$
$\frac{z_{3} - z_{1}}{z_{2} - z_{1}} = \frac{z_{2} - z_{1}}{z_{3} - z_{1}} [z_{3} - z_{1}] [z_{2} - z_{1}] = [z_{2} - z_{1}]^{2} o r {z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$ Now for the an equiliteral traingle , circumceral is the same as the centrid so that
$z_{0} = z_{1} + z_{2} + z_{3} = / 3 o r 9 {z_{0}}^{9} = {z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1} = 2 z_{1} z_{2} + 2 z_{2} z_{3} + 2 z_{3} z_{1}$
$= 3 z_{0} 2 = 3 z_{1} z_{2} + 3 z_{2} z_{3} + 3 z_{3} z_{1}$

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Q. Let the complex number

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Let the complex number z1,z2,z3 be the varticles of an equilateral triangle . let z0 be the cirumcental of the triangle . prove that z12+z22+z32=3z20

Let the complex number $z_{1}, z_{2}, z_{3}$ be the varticles of an equilateral triangle . let $z_{0}$ be the cirumcental of the triangle . prove that ${z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 z_{0}^{2}$