How do you find the cube root of z = 1 + i?

Firstly we express z in polar form:
 
z = Reⁱθ
 
where |z| = (Re² + Im²)^0.5 = (1² + 1²)^0.5 = 2^0.5
 
θ = arg z = tan^-1(Im/Re) = tan^-1(1/1) = π/4
 
Therefore z = (2^0.5)eⁱπ/4
 
We can add on any multiple of 2π to the argument of z without affecting the value of the complex number:
 
z = (2^0.5)eⁱ(π/4 + 2πn)
 
where n is an integer
 
We then take cube roots of both sides (not forgetting to cube root the modulus R as well as the exponent):
 
z^1/3 = (2^1/6)e^{i(π/12 + 2}πn/3) = (2^1/6)e^{i(π + 8}πn)/12
 
Because we are calculating the cube root, we expect three solutions. To find these three roots, we substitute in three consecutive integers into n. We will choose n = 0, 1, 2.
 
Solution 1 (with n=0): z^1/3 = (2^1/6)e^i(π/12)
Solution 2 (with n=1): z^1/3 = (2^1/6)e^i(3π/4)
Solution 3 (with n=2): z^1/3 = (2^1/6)e^i(17π/12)
 
We can convert these back into Cartesian form using:
 
z = R(cosθ + i sinθ)
 
We find that:
 
Solution 1: z^1/3 =(2^1/6)(cos(π/12) + i sin(π/12)) = 1.08 + 0.291i
Solution 2: z^1/3 = (2^1/6)(cos(3π/4) + i sin(3π/4)) = -0.794 +0.794i
Solution 3: z^1/3 = (2^1/6)*(cos(17π/12) + i sin(17π/12)) = -0.291-1.084i