if Z1 and Z2are 2 complex nos show that they can form an equilatrel triangle with origin if and only if (Z1/Z2 + Z2/Z1)=1

Dear student,

LET A,B,C BE THE VERTICES REPRESENTED BY Z1,Z2,Z3...ABC IS EQUILATERAL HENCE

AB=BC=CA=R SAY............
AB=Z2-Z1...|Z2-Z1|=R
BC=Z3-Z2...|Z3-Z2|=R
CA=Z1-Z3...|Z1-Z3|=R

ANGLES..ABC=BCA=CAB=60 DEG.
ANGLE BETWEEN AB & BC IS ANGLE ABC..THAT IS BETWEEN Z2-Z1 AND Z3-Z2..ETC,..

HENCE IF WE TAKE
AB=Z2-Z1=RE^(iT)............I..........WHERE T IS SOME ANGLE....THEN
BC=Z3-Z2= RE^i(T+60)...........II
CA=Z1-Z3=RE^i(T-60).....................III
SQUARING EQNS.I,II,III AND ADDING
(Z2-Z1)^2+(Z3-Z2)^2+(Z1-Z3)^2=R^2[E^(2iT)+E^2i(T+60)+E^2i(T-60)]
2[Z1^2+Z2^2+Z3^2-Z1Z2-Z2Z3-Z3Z1]=(R^2)(E^2iT)[1+E^(120i)+E^(-120i)]

[1+COS(120)+iSIN(120)+COS(-120)+iSIN(-120)]=(R^2)(E^2iT)[1-0.5+iSIN(120)-0.5-iSIN(120)]=0
Z1^2+Z2^2+Z3^2-Z1Z2-Z2Z3-Z3Z1=0
Z1^2+Z2^2+Z3^2=Z1Z2+Z2Z3+Z3Z1

Ultimately it gives (Z1/Z2 + Z2/Z1)=1

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