Q.266. To find fifth root of a given complex number.

Question 266.
Find the fifth roots of 32(cos300° + i sin300°). Express the roots in the form a + ib.

Answer 266.
According to De Moivre's theorem,
(cosθ + isinθ)^n = cosnθ + i sinnθ
Also cosθ = cos(2kπ+θ) and sinθ = sin(2kπ+θ)
=> (cosθ + isinθ)^n = cos n(2kπ+θ) + i sin n(2kπ+θ)
If nth root is desired,
(cosθ + i sinθ)^(1/n) = cos (2kπ/n + θ/n) + i sin n(2kπ/n + θ/n)

It follows from the above theory that
[32(cos300° + i sin300°)^(1/5)
= 2[cos(2kπ/5 + 60°) + i sin(2kπ/5 + 60°)], k = 0, 1, 2, 3, 4
5 values of k give 5 roots of the given function.
k = 0 => 2(cos60° + i sin60°) = 1 + i (√3/2)
k = 1 => 2(cos132° + i sin132°) ≈ - 1.338 + i (1.486)
k = 2 => 2(cos204° + i sin204°) ≈ - 1.827 - i (0.813)
k = 3 => 2(cos276° + i sin276°) ≈ 0.209 - i (1.989)
k = 4 => 2(cos348° + i sin348°) ≈ 1.956 - i (0.416).

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