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

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Firstly we express z in polar form:

 

z = R*ei*θ

 

where |z| = (Re2 + Im2)0.5 = (12 + 12)0.5 = 20.5

 

θ = arg z = tan-1(Im/Re) = tan-1(1/1) = π/4

 

Therefore z = (20.5)*ei*π/4

 

We can add on any multiple of 2π to the argument of z without affecting the value of the complex number:

 

z = (20.5)*ei*(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):

 

z1/3 = (21/6)*ei(π/12 + 2*π*n/3) = (21/6)*ei(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): z1/3 = (21/6)*ei(π/12)

Solution 2 (with n=1): z1/3 = (21/6)*ei(3π/4)

Solution 3 (with n=2): z1/3 = (21/6)*ei(17π/12)

 

We can convert these back into Cartesian form using:

 

z = R*(cosθ + i sinθ)

 

We find that:

 

Solution 1: z1/3 =(21/6)*(cos(π/12) + i sin(π/12)) = 1.08 + 0.291i

Solution 2: z1/3 = (21/6)*(cos(3π/4) + i sin(3π/4)) = -0.794 +0.794i

Solution 3: z1/3 = (21/6)*(cos(17π/12) + i sin(17π/12)) = -0.291-1.084i

Aldo E. GCSE Maths tutor, A Level Maths tutor, A Level Further Mathem...

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