SOLUTION: If z1+z2+z3=0 and z1^2+z2^2+z3^2=0, prove that |z1|=|z2|=|z3|

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: If z1+z2+z3=0 and z1^2+z2^2+z3^2=0, prove that |z1|=|z2|=|z3|       Log On


   



Question 888022: If z1+z2+z3=0 and z1^2+z2^2+z3^2=0, prove that |z1|=|z2|=|z3|

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
z%5B1%5D%2Bz%5B2%5D%2Bz%5B3%5D%22%22=%22%22%220%22

z%5B3%5D%22%22=%22%22-z%5B2%5D-z%5B1%5D

Substitute in

z%5B1%5D%5E2%2Bz%5B2%5D%5E2%2Bz%5B3%5D%5E2%22%22=%22%22%220%22

z%5B1%5D%5E2%2Bz%5B2%5D%5E2%2B%28-z%5B2%5D-z%5B1%5D%29%5E2%22%22=%22%22%220%22

z%5B1%5D%5E2%2Bz%5B2%5D%5E2%2B%28z%5B2%5D%5E2%2B2z%5B1%5Dz%5B2%5D%2Bz%5B1%5D%5E2%29%22%22=%22%22%220%22

z%5B1%5D%5E2%2Bz%5B2%5D%5E2%2Bz%5B2%5D%5E2%2B2z%5B1%5Dz%5B2%5D%2Bz%5B1%5D%5E2%22%22=%22%22%220%22

2z%5B1%5D%5E2%2B2z%5B2%5D%5E2%2B2z%5B1%5Dz%5B2%5D%22%22=%22%22%220%22

z%5B1%5D%5E2%2Bz%5B2%5D%5E2%2Bz%5B1%5Dz%5B2%5D%22%22=%22%22%220%22

z%5B2%5D%5E2%2Bz%5B1%5Dz%5B2%5D%2Bz%5B1%5D%5E2%22%22=%22%22%220%22

z%5B2%5D%22%22=%22%22%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

z%5B2%5D%22%22=%22%22%28-z%5B1%5D+%2B-+sqrt%28z%5B1%5D%5E2-4%2A1%2Az%5B1%5D%5E2+%29%29%2F%282%2A1%29+


z%5B2%5D%22%22=%22%22%28-z%5B1%5D+%2B-+sqrt%28z%5B1%5D%5E2-4z%5B1%5D%5E2+%29%29%2F2+

z%5B2%5D%22%22=%22%22%28-z%5B1%5D+%2B-+sqrt%28-3z%5B1%5D%5E2%29%29%2F2+

z%5B2%5D%22%22=%22%22%28-z%5B1%5D+%2B-+z%5B1%5D%2Ai%2Asqrt%283%29%29%2F2+

z%5B2%5D%22%22=%22%22z%5B1%5D%28-1+%2B-+i%2Asqrt%283%29%29%2F2+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A+abs%28+%28-1+%2B-+i%2Asqrt%283%29%29%2F2%29+%29+


abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A+abs%28+-1%2F2+%2B-+i%2Asqrt%283%29%2F2+%29+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A+sqrt%28%28-1%2F2%29%5E2+%2B%28%22%22+%2B-+sqrt%283%29%2F2%29%5E2+%29+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A+sqrt%281%2F4+%2B3%2F4+%29+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A+sqrt%281+%29+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+%2A%281+%29+

abs%28z%5B2%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+

by symmetry, we may swap z%5B2%5D and z%5B3%5D

abs%28z%5B3%5D%29%22%22=%22%22+abs%28z%5B1%5D%29+

abs%28z%5B1%5D%29%22%22=%22%22+abs%28z%5B2%5D%29+%22%22=%22%22+abs%28z%5B3%5D%29+

Edwin