The three vertices of a triangle are represented by the complex numbers, $0, z_{1} and z_{2}$ . If the triangle is equilateral, then show that $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$ . Further if $z_{0}$ is circumcentre then prove that $z_{1}^{2} + z_{2}^{2} = 3 z_{0}^{2}$

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Solution

To prove: $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$

$z_{1}^{2} + z_{2}^{2} = 3 z_{0}^{2}$

where $z_{0}$ is the circumcenter of the equilateral triangle.

Let this triangle be T

This is an equilateral triangle with each angle $\frac{π}{3}$ or $60^{o}$

Proof: We know that

$- z_{1}$ and $z_{2} - z_{1}$ are two sides of which meet at $z_{1}$

From the geometry of $T$ , it follows that $- z_{1}$ is at an angle of $π$ to $z_{2} - z_{1}$ .

Similarly, $z_{1} - z_{2}$ and $- z_{2}$ are two sides of T which meet at $z_{2}$ and $z_{1} - z_{2}$ is at angle of $\frac{π}{3}$ to $- z_{2}$

We know that,

$- z_{1} = e^{i \frac{π}{3}} (z_{2} - z_{1})$ .....(1)

$z_{1} - z_{2} = e^{i \frac{π}{3}} (- z_{2}) . . . . . (2)$

Then,

$\frac{- z_{1}}{z_{1} - z_{2}} = \frac{z_{2} - z_{1}}{- z_{2}}$ ${(1) d i v i d e n d b y (2)}$

$\Rightarrow (- z_{1}) (- z_{2}) = (z_{2} - z_{1}) (z_{1} - z_{2})$

$\Rightarrow z_{1} z_{2} = z_{1} z_{2} - z_{2}^{2} - z_{1}^{2} + z_{z} z_{2}$

$\Rightarrow z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$ ....(3) Here proved.

We know that,

circumcentre and centroid of an equilateral triangle coincide

$\Rightarrow$ $z_{0} = \frac{1}{3} (0 + z_{1} + z_{2})$

$\Rightarrow$ $3 z_{0} = z_{1} + z_{2}$

squaring both sides

${(3 z_{0})}^{2} = {(z_{1} + z_{2})}^{2}$

$\Rightarrow (3 z_{0})^{2} = z_{1}^{2} + z_{2}^{2} + 2 z_{1} z_{2}$ ...(4)

Substituting (3) in (4)

${(3 z_{0})}^{2} = z_{1}^{2} + z_{2}^{2} + 2 (z_{1}^{2} + z_{2}^{2})$

${(3 z_{0})}^{2} = z_{1}^{2} + z_{2}^{2} + 2 z_{1}^{2} + 2 z_{2}^{2}$

$\Rightarrow {3 (3 z_{0})}^{2} = 3 (z_{1}^{2} + z_{2}^{2})$

$\Rightarrow {(3 z_{0})}^{2} = (z_{1}^{2} + z_{2}^{2})$

Hence proved

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Similar questions

Q. 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}

Q. Let the complex number

z_{1}, z_{_{2}}, z_{3}

be the vertices of an equilateral triangle. Let z

z_{0}

be the circumcentre of the triangle , then proove that

{z_{1}}^{2} + {z_{2}}^{2} + {z_{3}}^{2} = 3 {z_{0}}^{2}

Q. Let

z_{1}, z_{2}, & z_{3}

be the vertices and

z_{0}

be the circumcentre of an equilateral triangle. then

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

Q. Show that the complex numbers

z_{1}, z_{2}

and the origin form an equilateral triangle only if

z_{1}^{2} + z_{2}^{2} - z_{1} z_{2} = 0

Let the complex numbers z₁,z₂and z₃ be the vertices of an equilateral triangle. Let z₀ be the circumcentre of the triangle, then $z_{1}^{2} + z_{2}^{2} + z_{3}^{2}$ =

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The three vertices of a triangle are represented by the complex numbers, 0,z1 and z2. If the triangle is equilateral, then show that z21+z22=z1z2. Further if z0 is circumcentre then prove that z21+z22=3z20

The three vertices of a triangle are represented by the complex numbers, $0, z_{1} and z_{2}$ . If the triangle is equilateral, then show that $z_{1}^{2} + z_{2}^{2} = z_{1} z_{2}$ . Further if $z_{0}$ is circumcentre then prove that $z_{1}^{2} + z_{2}^{2} = 3 z_{0}^{2}$