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

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Solution

Since $z_{1}, z_{2}, z_{3}$ are the vertices of an equilateral triangle
Therefore circumcenter $(z_{0}) =$ centroid $(\frac{z_{1} + z_{2} + z_{3}}{3})$ ...(1)
Also for equilateral triangle
$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}$ ...(2)
On squaring (1), we get
${9 z_{0}}^{2} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1} z_{2} + z_{2} z_{3} + z_{3} z_{1}) \Rightarrow {9 z_{0}}^{2} = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2 (z_{1}^{2} + z_{2}^{2} + z_{3}^{2}) \Rightarrow {3 z_{0}}^{2} = z_{1}^{2} + z_{2}^{2} + z_{3}$

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Let the complex number z1,z2, and z3 be the vertices of an equilateral triangle. Let z0 be the circumcentre of the triangle. Then show that z21+z22+z23=3z20.

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