Question

If $z_1+z_2+z_3=A,z_1+z_2\omega +z_3{\omega }^2=B,z_1+z_2{\omega }^2+z_3\omega =C$, find $z_1,z_2,z_3$ in terms of $A,B,C,\omega$

Answer 1

$z_1+z_2+z_3=A$ ___ (1)

$z_1+z_2\omega +z_3{\omega }^2=B$ ___ (2)

$z_1+z_2{\omega }^2+z_3\omega =c$ ___ (3)

$\left(1\right)+\left(2\right)+\left(3\right)$

$3z_1+z_2(1+\omega +{\omega }^2)+z_3(1+\omega +{\omega }^2)=A+B+C$

$z_1=\frac{A+B+C}{3}$

$\left(1\right)\times \omega +\left(2\right)\times {\omega }^2+\left(3\right)\times {\omega }^3$

$z_1(1+\omega +{\omega }^2)+z_2(\omega +{\omega }^3+{\omega }^2)+z_3(\omega +{\omega }^4+{\omega }^4)=A\omega +B{\omega }^2+C{\omega }^3$

$z_3(\omega +2{\omega }^{\prime })=A\omega +B{\omega }^2+C{\omega }^3$

$z_3=\frac{A\omega +B{\omega }^2+C}{3\omega }=\frac{A\omega +B{\omega }^2+C}{3\omega }$

$z_2=A-z_1-z_3=A-\frac{A+B+C}{3}-\frac{A\omega +B{\omega }^2+C}{3\omega }$

$=\frac{2A-B-C}{3}-\frac{A\omega +B{\omega }^2+C}{3\omega }$

$z_2=\frac{A\omega -B\omega -B{\omega }^2-C\omega -C}{3\omega }$