a+ bi can be written as \(\displaystyle re^{i\theta}\) where \(\displaystyle r= \sqrt{a^2+ b^2}\) (the positive root) and \(\displaystyle \theta= arctan\left(\frac{b}{a}\right)\). That is the "mod-arg form" Harry the cat refers to. In that form, \(\displaystyle \left(r e^{i\theta}\right)^{1/2}= r^{1/2}e^{i\theta/2}\).
If you are not familiar with that form this can be done more fundamentally, though harder, by writing the root as a+ bi and the \(\displaystyle (a+ bi)^2= (2- 2i\sqrt{3})^2\) so \(\displaystyle a^2- b^2+ 2abi= -8- 8\sqrt{3}i\) and then \(\displaystyle a^2- b^2= -8\) and \(\displaystyle 2ab= -8\sqrt{3}\). Solve those two equations for a and b.
