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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