What is the cube root of -2 plus 2i?

Answer 1

Step 1. Convert (-2 + 2i) into polar form:

r = √((-2)2 + 22) = √8 = 23/2

tan θ = 2/-2 = -1

⇒ θ = arctan(-1) = 3π/4 (in 2nd quadrant)

⇒ (-2 + 2i) = 23/2(cos 3π/4 + i sin 3π/4)

Step 2. Apply DeMoivre's Theorem:

3√(-2 + 2i) = (-2 + 2i)1/3

= (23/2(cos 3π/4 + i sin 3π/4))1/3

= 21/2(cos π/4 + i sin π/4)

Which provides one root; for the other two roots, add 1/3 a turn = 2π/3 and 2/3 a turn = 4π/3 to the angle:

z1 = √2(cos π/4 + i sin π/4)

z2 = √2(cos 11π/12 + i sin 11π/12)

z3 = √2(cos 19π/12 + i sin 19π/12)

[adding the next third of a turn comes back to the first root.]