x,y,z are non zero real numbers. a,b,c are complex numbers such that |a| = |b| = |c| =1 .
If x+y+z = 0 = ax + by + cz , how do I prove a=b=c ?

Question

x,y,z are non zero real numbers. a,b,c are complex numbers such that |a| = |b| = |c| =1 .
If x+y+z = 0 = ax + by + cz , how do I prove a=b=c ?

experimentx · Answer

$$ (x+y)^2 = z^2 $$
complex numbers add up like vectors, write $ a = \cos \theta_1 + i \sin \theta_1, $ ... and so on.
$$ ax+by+cz = 0 \
ax+by = -cz \ 
(x \cos \theta_1  + y \cos \theta_2 ) + i (x \sin \theta_1  + y \sin \theta_2 ) = z( \cos \theta_3 + i \sin \theta_3) \ 
|(x \cos \theta_1  + y \cos \theta_2 ) + i (x \sin \theta_1  + y \sin \theta_2 )| = |z( \cos \theta_3 + i \sin \theta_3)| \
x^2 + 2xy ( \cos \theta_1 \cos \theta_2 +\sin \theta_1 \sin \theta_2) + y^2 = z^2 \
x^2 + 2 xy \cos(\theta_1 - \theta_2) +y^2 = z^2 = (x+y)^2$$

experimentx · Answer

that implies $\theta_1 = \theta_2$ next time we will switch variables and prove $\theta_1 = \theta_3$ ... hence the argument of a,b,c are equal

shubhamsrg · Answer

yes that was easy and i should have been able to do that hmm
thanks again sire.

experimentx · Answer

yw