Find two square roots for the following complex number. Write your answers in standard form, separated by commas.
\[
2-2 \sqrt{3} i
\]

Step 1 ：First, we write the complex number in polar form. The magnitude of the complex number is given by \( r = \sqrt{2^2 + (-2\sqrt{3})^2} = 4 \). The argument of the complex number is given by \( \theta = \arctan\left(\frac{-2\sqrt{3}}{2}\right) = -\frac{\pi}{3} \). So, the complex number in polar form is \( 4(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) \).

Step 2 ：Next, we find the square roots of the complex number. The magnitude of the square roots is \( \sqrt{4} = 2 \). The arguments of the square roots are \( \frac{-\frac{\pi}{3}}{2} = -\frac{\pi}{6} \) and \( \frac{-\frac{\pi}{3} + 2\pi}{2} = \frac{5\pi}{6} \).

Step 3 ：So, the square roots of the complex number in polar form are \( 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) \) and \( 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6})) \).

Step 4 ：Finally, we convert these back to standard form. The first square root is \( 2\cos(-\frac{\pi}{6}) + 2i\sin(-\frac{\pi}{6}) = \sqrt{3} + i \). The second square root is \( 2\cos(\frac{5\pi}{6}) + 2i\sin(\frac{5\pi}{6}) = -\sqrt{3} - i \).

Step 5 ：So, the two square roots of the complex number \( 2 - 2\sqrt{3}i \) are \( \boxed{\sqrt{3} + i} \) and \( \boxed{-\sqrt{3} - i} \).