Find the cube roots of 27(cos 330° + i sin 330°).?

Hints: De Moivre's formula; exponential notation; Euler's formula.

3(cos 110 + i sin 110)
and
3(cos 230 + i sin 230)
and
3(cos 350 + i sin 350).

I divided the angle by 3, and then looked for the two other angles that would be equally spaced around a circle.

The nth roots of r(cos(A°) + i sin(A°)) are ⁿ√r (cos((A°+ k×360°)/n) + i sin((A°+ k×360°)/n)) for k = 0 to n-1.
∛27 = 3 and (330°+ k×360°)/3 is 110°, 230°, and 350°
So the cube roots are 3(cos 110° + i sin 110°), 3(cos 230° + i sin 230°) and 3(cos 350° + i sin 350°)

Graph: https://www.desmos.com/calculator/ol1fwzxwev

Z = 27.[cos(330) + i.sin(330)]

Then you must find z, such as: z³ = Z


The modulus of Z is 27, so the modulus of z is: 27^(1/3) = [3^(3)]^(1/3) = 3^[3 * (1/3)] = 3^(1) = 3

The argument Z is 330, so the argument of the first root is: (330/3) = 110 then you can deduce that the first root is:

z₁ = 3.[cos(110) + i.sin(110)] → and you add (360/3), i.e. (120) to obtain the second root

z₂ = 3.[cos(110 + 120) + i.sin(110 + 120)] → and you add 120 once again to obtain the third root

z₃ = 3.[cos(110 + 120 + 120) + i.sin(110 + 120 + 120)]


It gives us:

z₁ = 3.[cos(110) + i.sin(110)]

z₂ = 3.[cos(230) + i.sin(230)]

z₃ = 3.[cos(350) + i.sin(350)]

3(cos 110° + i sin 110°), 3(cos 230° + i sin 230°), 3(cos 350° + i sin 350°)

The cube root of 27 = 3 because 3*3*3 = 27
3*110 = 330, 3*230 = 690 = 360+330, 3*350 = 720+330

k = 0,1,2. In this case Θ = 330° and n = 3
(27^1/n)*[cos(Θ+ k*360°)/n + i*sin(Θ+k*360)/n]

z³ = 27cis330° = 27cis(11π/6) = 27e^[(11π/6 + 2kπ)i]
z = 3e^[(11π/18 + 2kπ/3)i]
z₀ = 3e^(11πi/18) = 3cis110°
z₁ = 3e^(23πi/18) = 3cis230°
z₂ = 3e^(35πi/18) = 3cis350°


Ans:
z₀ = 3(cos110° + i sin110°) ≈ -1.0261 + 2.8191 i
z₁ = 3(cos230° + i sin230°) ≈ -1.9284 - 2.2981 i
z₂ = 3(cos350° + i sin350°) ≈ 2.9544 - 0.5209 i