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Question

If z1,z2,z3 are the vertices of equilateral triangle with centroid z0, then z21+z22+z23 equals

A
z20
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B
2z20
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C
3z20
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D
9z20
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Solution

The correct option is B 3z20
If z0 is the centroid of the triangle then z0=z1+z2+z33
3z0=z1+z2+z3
To find z21+z22+z23
we have (a+b+c)2=a2+b2+c2+2(ab+bc+ca)
or a2+b2+c2=(a+b+c)22(ab+bc+ca)
z21+z22+z23=(z1+z2+z3)22(z1z2+z2z3+z3z1)
z21+z22+z23=(3z0)22(z1z2+z2z3+z3z1)
z21+z22+z23=9z022(z1z2+z2z3+z3z1) ..........(1)
In an equilateral triangle all the sides are equal.
AB=BC=CA=a
ABC=π3
From coni method,
z1z2z3z2=aaeiπ3 ....(2)
and BAC=π3
From coni method,
z3z1z2z1=aaeiπ3 ....(3)
From equations (2) and (3), we get
z1z2z3z2=z3z1z2z1
(z1z2)(z1z2)=(z3z2)(z3z1)
z21+z22+z23=z1z2+z2z3+z3z1 on simplification
Substituting for z1z2+z2z3+z3z1=z21+z22+z23 in eqn(1) we get
z21+z22+z23=9z022(z1z2+z2z3+z3z1)
z21+z22+z23=9z022(z21+z22+z23)
3(z21+z22+z23)=9z02
z21+z22+z23=3z02


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