Cube roots of unity:
For n = 3, we obtain the cube roots of unity and they are +They are usually shown by 1, ω and ω2 and geometrically shown by the vertices of an equilateral triangle whose circumradius is unity and circumcentre is origin.
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Note:
- ω3 = 1 and 1 + ω + ω2 = 0
- It can be easily proved that
1 + ωn + ω2n = 3 (n is a multiple of 3)
1 + ωn + ω2n = 0 (n is an natural number, not a multiple of 3)
Problem: Provided z1 + z2 + z3 = A ,
z1 + z2w + z3w2 = B
z1 + z2w2 + z3w = C
where w is cube root of unity.
Express z1, z2, z3 in term of A, B , C.
Solution: Adding three condition
z1 = A+B+C/3
Multiplying z1 + z2 + z3 = A, z1 + z2w + z3w2 = B, z1 + z2w2 + z2w = C by 1 , w2 ,w and adding we obtain
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