Complex NumbersHard
Question
If p = a + bω + cω2; q = b + cω + aω2 and r = c + aω + bω2 where a, b, c ≠ 0 and w is the non-real complex cube root of unity, then :
Options
A.p + q + r = a + b + c
B.p2 + q2 + r2 = a2 + b2 + c2
C.p2 + q2 + r2 = 2(pq + qr + rp)
D.none of these
Solution
p + q + r = (a + b + c) (1 + ω + ω2) = 0
(p + q + r)2 = 0
p2 + q2 + r2 + 2pq + 2qr + 2 rp = 0
p2 + q2 + r2 = (a2 + b2 + c2) (1 + ω + ω2) + (ab + bc + ca) (1 + ω + ω2) = 0
pq + qr + rp = 0
(p + q + r)2 = 0
p2 + q2 + r2 + 2pq + 2qr + 2 rp = 0
p2 + q2 + r2 = (a2 + b2 + c2) (1 + ω + ω2) + (ab + bc + ca) (1 + ω + ω2) = 0
pq + qr + rp = 0
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