Answer
$Two\,\,square\,\,roots\,\,of\,\,-2 + 2 \sqrt{3} i$
$\sqrt[2]{-2 + 2 \sqrt{3} i}=1+\sqrt{3}i \\
\sqrt[2]{-2 + 2 \sqrt{3} i}=-1-\sqrt{3}i\\
$
Work Step by Step
$to\,\,find\,\,nth\,root\,of\,a\,complex\,number\,z=r(cos\theta+isin\theta) :\\
\sqrt[n]{r}\left(\cos\left(\frac{\theta+2\pi k}{n}\right)+i\sin\left(\frac{\theta+2\pi k}{n}\right)\right)\\
k=0,1,2,.......,n-1 \\so\,\,
-2 + 2 \sqrt{3} i=4 \cos{\left (\frac{2\pi}{3} \right )} + 4 i \sin{\left (\frac{2\pi}{3} \right )}\\
r=4\,\,,\theta= \frac{2\pi}{3}\,\,,n=2 \\
for\,k=0\\
\sqrt[2]{4}\left(\cos\left(\frac{\left(\frac{2\pi}{3}\right)+2\pi \cdot 0}{2}\right)+i\sin\left(\frac{\left(\frac{2\pi}{3}\right)+2\pi \cdot 0}{2}\right)\right)\\=2\left(\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)=1+\sqrt{3}i \\
for\,k=1\\
\sqrt[2]{4}\left(\cos\left(\frac{\left(\frac{2\pi}{3}\right)+2\pi \cdot 1}{2}\right)+i\sin\left(\frac{\left(\frac{2\pi}{3}\right)+2\pi \cdot 1}{2}\right)\right)\\=2\left(\cos\left(\frac{4 \pi}{3}\right)+i\sin\left(\frac{4 \pi}{3}\right)\right)=-1- \sqrt{3}i \\
so\,\,\\
\sqrt[2]{-2 + 2 \sqrt{3} i}=1+\sqrt{3}i \\
\sqrt[2]{-2 + 2 \sqrt{3} i}=-1-\sqrt{3}i\\
$
