This one IS a little difficult!!
The "formula" for finding roots of complex numbers is given by (r)^(1/N)*e^( i (Theta/N + (2 * pi * k) / N )
where
r = SQRT(a^2 + b^2) for the complex number a + bi = in our case, 0 - 8i....so r = SQRT(0*2 + 8*2) = SQRT(64) = 8
N = root we are trying to take....in this case, the 3rd one.....so, (r)^(1/3) = (8)*(1/3) = 2
Note that e^( i (Theta/N + (2 * pi * k) / N ) can be written as cos( (Theta/N + (2 * pi * k) / N )) + i * sin((Theta/N + (2 * pi * k) / N ) where k = 0, 1, 2, etc.
(Stick with me...it'll make sense in a minute!!)
We need to find THETA.....This is easy.... 0 - 8i is just a point on the "i" axis in the complex plane 8 units "down" from (0,0)..i.e., -(pi/2)
Now, we're ready to find the roots!!
Let k = 0 for the first root
So we have.... 2 * (cos(( -pi/2 / 3) + 2 * Pi * 0 / 3) + i * sin (( -pi/2 / 3) + 2 * Pi * 0 / 3))
= 2 (cos( -pi/6) + i * sin( -pi / 6)) = 2 ((SQRT(3) / 2) + i (-1/2)) = SQRT(3) - i
If you cube SQRT(3) - i, I think you'll find that you get 0 -8i !!!
Now, for the next root This time, let k =1
So we have.... 2 * (cos( -pi/2 / 3) + 2 * Pi * 1 / 3) + i * sin (( -pi/2 / 3) + 2 * Pi * 1 / 3))
= 2 *(cos(-pi/6 + 2 * pi * 1 / 3) + i * sin (-pi/6 + 2 * pi * 1 / 3))
= 2(cos(1//2) * pi) + i * sin ((1/2) * pi)) = 2(0 - i) = 2i
Again, if you cube this, you should get 0 - 8i
Almost finished!! Let's take the LAST root!! You guessed it, let k =2 this time
So we have ........2 * (cos( -pi/2 / 3) + 2 * Pi * 2 / 3) + i * sin (( -pi/2 / 3) + 2 * Pi * 2 / 3))
= 2* (cos (7/6) * pi) + i* sin (7/6)* pi))
= 2 (-SQRT(3)/2) + i *(-1/2)
= - (SQRT(3) + i)
Again, if you cube this, you'll get 0 - 8i
Whew!! THAT was a LOT of work, huh?? At least the reference angles were easy to find. (It's sometimes WAY messier!!)
Hope this helps. (I hope I got all my parenthesis in the right places, too!!)
BTW...here's a good website to review complex numbers...it's pretty detailed!!
http://tutorial.math.lamar.edu/Extras/ComplexPrimer/ComplexNumbers.aspx
+129852 
