Q.431. Root of a Complex Number - De Moivre's Theorem

Question 431.
Z^10 = -1 (Z is a complex number).
Prove that the answers Z1, Z2, Z3....Z10 are Geometrical progression.

Answer 431.
Z^10 = -1
=> Z
= (-1)^(1/10)
= (cosπ + isinπ)^(1/10)
= [cos(2k+1)π + i sin(2k+1)π]^(1/10)
= cos (2k+1) (π/10) + i sin(2k+1) (π/10)
k = 10, 1, 2, 3, 4, .... 9 give the values of Z1, Z2, Z3, ... Z10
For them to be in G.P., we need to prove that the ratio of two successive roots is the same

Let Zn and Zn+1 be two successive roots
=> Zn = cos(2n-1) (π/10) + i sin(2n-1) (π/10)
and Zn+1 = cos(2n+1) (π/10) + i sin(2n+1) (π/10)
=> Zn+1 / Zn
= [cos(2n+1) (π/10) + i sin(2n+1) (π/10)] / [cos(2n-1) (π/10) + i sin(2n-1) (π/10)]
= cos(π/5) + isin(π/5)
As this is independent of n, the ratio is the same for any two successive terms
=> Z1, Z2, Z3, ... Z10 are in G.P.

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