Let z = 2 - 2√3 i. In polar form, we have
[tex]z = |z| e^{i\arg(z)}[/tex]
where
[tex]|z| = |2-2\sqrt3| = \sqrt{2^2 + (-2\sqrt3)^2} = 4[/tex]
[tex]\arg(z) = \tan^{-1}\left(-\dfrac{2\sqrt3}2\right) = -\tan^{-1}(\sqrt3) = -\dfrac\pi3[/tex]
Equivalently,
[tex]z = 4\left(\cos\left(-\dfrac\pi3\right) + i\sin\left(-\dfrac\pi3\right)\right)[/tex]
Let w be a complex number such that w ² = z. By DeMoivre's theorem,
[tex]z = 4\left(\cos\left(-\dfrac\pi3\right) + i\sin\left(-\dfrac\pi3\right)\right) \\\\ \implies z^{1/2} = 4^{1/2} \left(\cos\left(\dfrac{-\frac\pi3+2k\pi}{2}\right) + i \sin\left(\dfrac{-\frac\pi3+2k\pi}{2}\right)\right)[/tex]
where k ∈ {0, 1}. So the two square roots of z are
[tex]z^{1/2} = \begin{cases}2\left(\cos\left(-\dfrac\pi6\right) + i\sin\left(-\dfrac\pi6\right)\right) & \text{for }k=0 \\\\ 2\left(\cos\left(\dfrac{5\pi}6\right) + i\sin\left(\dfrac{5\pi}6\right)\right) & \text{for }k=1\end{cases}[/tex]
[tex]z^{1/2} = \begin{cases}2\left(\dfrac{\sqrt3}2 - \dfrac12 i\right) & \text{for }k=0 \\\\ 2\left(-\dfrac{\sqrt3}2 + \dfrac12 i\right) & \text{for }k=1\end{cases}[/tex]
[tex]\boxed{z^{1/2} = \begin{cases}\sqrt3 - i & \text{for }k=0 \\ -\sqrt3 + i & \text{for }k=1\end{cases}}[/tex]
