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
 

















Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
Win exciting gifts by                                                      answering the     questions    on            Discussion        Forum.    So      help              discuss         any                query   on        askiitians      forum    and         become  an       Elite            Expert      League                askiitian.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..
 
Askiitians Expert
Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com