How to find the cube roots of -2 2i in Cartesian form??

Question

...giving the real and imaginary parts of each root correct to three decimal places.

I'm not sure how to deal with this questions, some pointers plz?

Answer 1

The modulus of - 2 + 2i is sqrt(2^2 + 2^2) = 2 sqrt(2) = 2^(3 / 2)

- 2 + 2i = [ 2^(3 / 2) ] ( cos(t) + i sin(t) )

where

cos(t) = - 1 / sqrt(2) ...(1)

sin(t) = 1 / sqrt(2) ...(2)

t is a 2nd quadrant angle.

Using de Moivre's Theorem, the cube roots are:

[ 2^(1 / 2) ] ( cos(t / 3) + i sin(t / 3) ) ...(3)

for each distinct t satisfying both (1) and (2).

The solutions of (1) and (2) are:

t = 3pi / 4, 11pi / 4, 19pi / 4

giving

t / 3 = pi / 4, 11pi / 12, 19pi / 12.

Using these angles in (3) and evaluating, the roots are:

sqrt(2)(cos(pi / 4) + i sin(pi / 4))

= 1.000 + 1.000i

sqrt(2)(cos(11pi / 12 + i sin(11pi / 12))

= - 1.366 + 0.366i

sqrt(2)(cos(19pi / 12) + i sin(19pi / 12))

= 0.366 - 1.366i.