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

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

Firstly we express z in polar form:

z = R*e^i*θ

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)*e^i*π/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)*e^{i*(π/4 + 2*π*n)}

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*π*n/3)} = (2^1/6)*e^{i(π + 8*π*n)/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

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