Originally Posted by
HallsofIvy Well, first your values for u are wrong. $\displaystyle u= -1+ i 3^{1/2}$ and $\displaystyle u= -1- i 3^{1/2}$. You forgot the "i".
Now use the fact that $\displaystyle u= z^2$.
$\displaystyle z^2= -1+ i3^{1/2}$ so $\displaystyle z= \pm\sqrt{-1+ i 3^{1/2}}$
$\displaystyle z^2= -1- i3^{1/2}$ so $\displaystyle z= \pm\sqrt{-1- i3^{1/2}}$.
I presume you will want to use d'Moivre's formula to take those square roots. Since $\displaystyle 1^2+ (3^{1/2})^2= 4$ and $\displaystyle \frac{3^{1/2}}{-1}= -\sqrt{3}= arctan(\pi/3}$, in polar form, $\displaystyle -1+ i3^{1/2}= 2(cos(\pi/3)+ i sin(\pi/3))= 2e^{i\pi/3}$ and $\displaystyle -1- i3^{1/2}= 2(cos(\pi/3)- i sin(\pi/3))= 2e^{-i\pi/3}$.