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Question: If \(z_{1},z_{2},z_{3}\) are the vertices of an equilateral triangle with \(z_{0}\) as its circumcen...

If z1,z2,z3z_{1},z_{2},z_{3} are the vertices of an equilateral triangle with z0z_{0} as its circumcentre then changing origin to z0z_{0}, then (where z1,z2,z3z_{1},z_{2},z_{3} are new complex numbers of the vertices)

A

z12+z22+z32=0z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0

B

z1z2+z2z3+z3z1=0z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1} = 0

C

Both (1) and (2)

D

None of these

Answer

z12+z22+z32=0z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 0

Explanation

Solution

Sol. In an equilateral triangle the circumcentre and the centroid are the same point. So,

z0=z1+z2+z33\mathbf{z}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{z}_{\mathbf{1}}\mathbf{+}\mathbf{z}_{\mathbf{2}}\mathbf{+}\mathbf{z}_{\mathbf{3}}}{\mathbf{3}}z1+z2+z3=3z0z_{1} + z_{2} + z_{3} = 3z_{0} ..... (i)

To shift the origin at z0,z_{0}, we have to replace z1,z2,z3z_{1},z_{2},z_{3} and z0z_{0} by z1+z0,z2+z0,z3+z0z_{1} + z_{0},z_{2} + z_{0},z_{3} + z_{0}and 0+z00 + z_{0} then equation (i) becomes (z1+z0)+(z2+z0)+(z3+z0)=3(0+z0)(z_{1} + z_{0}) + (z_{2} + z_{0}) + (z_{3} + z_{0}) = 3(0 + z_{0})z1+z2+z3=0z_{1} + z_{2} + z_{3} = 0

On squaring z12+z22+z32+2(z1z2+z2z3+z3z1)=0z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + 2(z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1}) = 0

But triangle with vertices z1,z2z_{1},z_{2} and z3z_{3} is equilateral, then z12+z22+z32=z1z2+z2z3+z3z1z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = z_{1}z_{2} + z_{2}z_{3} + z_{3}z_{1} .....(iii)

From (ii) and (iii) we get, 3(Z12+Z22+Z32)=03(Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2}) = 0. Therefore, Z12+Z22+Z32=0.Z_{1}^{2} + Z_{2}^{2} + Z_{3}^{2} = 0.