First write it in polar form, or trigonometric form, however you call it, and after that use de moivre's formula to find its roots.
let z be a complex nr.
z=a+bi, writing it in polar forms : [tex]a=r cos(\theta),b=\ro sin\theta[/tex]
So,
[tex]z=r (cos\theta+isin\theta)[/tex]
now
[tex]z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos\frac{\theta +2k\pi}{n}+isin{\frac{\theta+2k\pi}{n})[/tex]
Now all you need to do is figure out what [tex]\theta[/tex] is, and your fine.
Or if you want the exponential representation of a complex nr:
[tex]e^{ix}=cosx+isinx[/tex]
[tex]e^{i\frac{\pi}{2}}=i[/tex] so we get
[tex]i=cos\frac{\pi}{2}+isin\frac{\pi}{2}[/tex]