Given: $z_{1} + z_{2} + z_{3} = A; z_{1} + z_{2} w + z_{3} w^{2} = B; z_{1} + z_{2} w^{2} + z_{3} w = C$ where $w$ is cube root of unity
Express $z_{1}, z_{2}, z_{3}$ in terms of $A, B, C$

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Solution

(e) We are given
$z_{1} + z_{2} + z_{3} = A . . . . (1)$
$z_{1} + z_{2} ω + z_{3} ω^{2} = B . . . . (2)$
$z_{1} + z_{2} ω^{2} + z_{3} ω = C . . . . (3)$
(a) Adding (1) and (2) and (3) we get
$3 z_{1} + z_{2} (1 + ω + ω^{2}) + z_{3} (1 + ω^{2} + ω) = A + B + C$
or $z_{1} = \frac{A + B + C}{3} [∵ 1 + ω + ω^{2} = 0]$
Now multiplying (1), (2) and (3) by $1, ω, ω$ respectively and adding we get
$z_{1} (1 + ω + ω^{2}) + z_{2} (1 + ω^{3} + ω^{3}) + z_{3} (1 + ω^{4} + ω^{2}) = A + B ω^{2} + C ω o r z_{2} = \frac{A + B ω^{2} + C ω}{3} [∵ 1 + ω^{4} + ω^{2} = 1 + ω + ω^{2} = 0; ω^{3} = 1]$
Similarly, $z_{3} = \frac{A + B ω + C ω^{2}}{3}$
This proves (e)

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