Question

if Z1, Z2 ,Z3 are vertices of equilateral triangle then prove that Z1^2 + Z2^2 + Z3^2 = Z1.Z2 +Z2.Z3+ Z3.Z1.

If Zo is circumcenter of triangle then also prove that 3Zo^2 = Z1^2 + Z2^2 + Z3^2​

Answer 1

Answer:

Since z1, z2, z3, are vertical of an equilateral triangle

therefore circcumcenter (z0) =centroid

(z1+z2+z3) /3...........1)

also for an equilateral triangle

z²1+z²2+z²3=z1z2+z2z3+z3z1........ 2)

on squaring (1). we get,

9z²0=z²1+z²2+z²3+2(z1z2+z2z3+z3z1)

9z²0=z²1+z²2+z²3+2(z²1+z²2+z²3)

3z²0=z²1+z²2+z²3

Answer 2

$\huge {\underline {\underline{ \sf {ANSWER \: : }}}}$

GIVEN :

$\star \: \sf{Z_{1} ,Z_{2} ,Z_{3} \: are \: vertices \: of \:an \: equilateral \: triangle.}$

TO PROVE :

$\star\: \: \sf{Z_{1}}^{2} + {Z_{2}}^{2} + {Z_{3}}^{2} = Z_{1}.Z_{2} + Z_{2}. Z_{3} + Z_{3}.Z_{1}$

$\star \: \: \sf{ {3Z_{o}}^{2} = \: Z_{1}}^{2} + {Z_{2}}^{2} + {Z_{3}}^{2}$

SOLUTION :

• Let us consider $\sf{Z_{1} ,Z_{2} ,Z_{3}}$ to be the vertices of the equilateral triangle.

We know that, each angle of the equilateral triangle is 60° i.e, $\sf \frac{ \pi}{3}$.

We have the formula, when the complex number is not rotating about its origin which is,

$\longrightarrow \: \boxed{ \sf{ \:\blue{\frac{Z_{3} -Z_{1}} { \mid \:{Z_{3} -Z_{1}} \mid} =\frac{Z_{2} -Z_{1}} { \mid \:{Z_{2} -Z_{1}} \mid \: \: } {e}^{i \alpha } }}}$

For, Anticlockwise Direction

$\implies \: \sf{ \: \frac{Z_{3} -Z_{1}} { \mid \:{Z_{3} -Z_{1}} \mid} =\frac{Z_{2} -Z_{1}} { \mid \:{Z_{2} -Z_{1}} \mid \: \: } {e}^{i \frac{ \pi}{3} } }$

As, there the triangle is equilateral, therefore $\sf \mid \:{Z_{3} - Z_{1}} \mid = \mid \:{Z_{2} - Z_{1}} \mid = l(lenght \: of\: the\: sides)$

$\implies \: \sf{ \: \frac{Z_{3} -Z_{1}} { \cancel{l}} =\frac{Z_{2} -Z_{1}} { \cancel{l}} {e}^{i \frac{ \pi}{3} } }$

$\implies \: \sf{ \: {Z_{3} -Z_{1}} = (\: Z_{2} -Z_{1} \:) {e}^{i \frac{ \pi}{3}}} \: \: \: \: \: \longrightarrow \: (1)$

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Now, for Clockwise Direction

$\implies \: \sf{ \: \frac{Z_{3} -Z_{2}} { \mid \:{Z_{3} -Z_{2}} \mid} =\frac{Z_{1} -Z_{2}} { \mid \:{Z_{1} -Z_{2}} \mid \: \: } {e}^{ - i \frac{ \pi}{3}}}$

Here also, $\sf \mid \:{Z_{3} - Z_{1}} \mid = \mid \:{Z_{2} - Z_{1}} \mid = l(lenght \: of\: the\: sides)$

$\implies \: \sf{ \: \frac{Z_{3} -Z_{2}} { \cancel{l}} =\frac{Z_{1} -Z_{2}} { \cancel{l}} {e}^{ - i \frac{ \pi}{3} } }$

$\implies \: \sf{ \: {Z_{3} -Z_{2}} = (\: Z_{1} -Z_{2} \:) {e}^{ - i \frac{ \pi}{3}}} \: \: \: \: \: \longrightarrow \: (2)$

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Now, Multiplying (1) and (2) we get,

$\implies \: \sf{ \:( {Z_{3} -Z_{1}})({Z_{3} -Z_{2}}) = (\: Z_{2} -Z_{1} \:).(\: Z_{1} -Z_{2} \:) \: {e}^{i \frac{ \pi}{3}}. {e}^{ - i \frac{ \pi}{3}}}$

$\implies \: \sf{ \:( {Z_{3} -Z_{1}})({Z_{3} -Z_{2}}) = (\: Z_{2} -Z_{1} \:).(\: Z_{1} -Z_{2} \:) \: {e}^{i \frac{ \pi}{3} - i \frac{ \pi}{3} } } \:$

$\implies \: \sf{ \:( {Z_{3} -Z_{1}})({Z_{3} -Z_{2}}) = - {(\: Z_{1} -Z_{2} )}^{2} \: {e}^{0}}$

$\sf \: \implies \: {Z_{3}}^{2} - Z_{2}.Z_{3} - Z_{1}.Z_{3} + Z_{1}.Z_{2} = - {Z_{1}}^{2} - {Z_{2}}^{2} + 2Z_{1}.Z_{2}$

$\sf \: \: \boxed { \: \green{ \therefore \: \: \sf {Z_{1}}^{2} + {Z_{2}}^{2} + \: {Z_{3}}^{2} = Z_{1}.Z_{2} + Z_{2}.Z_{3} + Z_{3}.Z_{1}}}$

Hence Proved .

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Now, For Equilateral Triangle,

As $Z_{o}$ is the Circumcentre of the triangle and also the centroid as the triangle is equilateral.

$\sf \therefore Z_{o} = \frac{Z_{1} + Z_{2} + Z_{3} }{3}$

$\sf \implies 3Z_{o} = Z_{1} + Z_{2} + Z_{3}$

Squaring both sides we get,

$\sf \: \implies {(3Z_{o}) }^{2} = ({Z_{1} + Z_{2} + Z_{3})}^{2}$

$\sf \implies {9Z_{o} }^{2} = {Z_{1} } ^{2} + {Z_{2}}^{2} + \: {Z_{3}}^{2} + 2Z_{1}.Z_{2} + 2Z_{2}.Z_{3} + 2Z_{3}.Z_{1}$

(Replacing, $\sf 2Z_{1}.Z_{2} + 2Z_{2}.Z_{3} + 2Z_{3}.Z_{1}$ by $\sf {Z_{1}}^{2} + {Z_{2}}^{2} + \: {Z_{3}}^{2}$, as we proved earlier we get)

$\sf \implies {9Z_{o} }^{2} = 3{Z_{1} } ^{2} + 3{Z_{2}}^{2} + \: 3{Z_{3}}^{2}$

$\sf \implies {3Z_{o} }^{2} = {Z_{1} } ^{2} + {Z_{2}}^{2} + \: {Z_{3}}^{2}$

$\sf {\therefore {\boxed {\green {{3Z_{o} }^{2} = {Z_{1} } ^{2} + {Z_{2}}^{2} + \: {Z_{3}}^{2} } } }}$

Hence Proved.