Answer
$two\,\,square\,\,root\,\,of\,\,2 + 2 \sqrt{3} i$
$\sqrt[2]{2 + 2 \sqrt{3} i}=\sqrt{3}+i \\
\sqrt[2]{2 + 2 \sqrt{3} i}=-\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)\\
2 + 2 \sqrt{3} i=4 \cos{\left (\frac{\pi}{3} \right )} + 4 i \sin{\left (\frac{\pi}{3} \right )}\\
r=4\,\,,\theta= \frac{\pi}{3}\,\,,n=2 \\
for\,k=0\\
\sqrt[2]{4}\left(\cos\left(\frac{\left(\frac{\pi}{3}\right)+2\pi \cdot 0}{2}\right)+i\sin\left(\frac{\left(\frac{\pi}{3}\right)+2\pi \cdot 0}{2}\right)\right)\\=2\left(\cos\left(\frac{\pi}{6}\right)+i\sin\left(\frac{\pi}{6}\right)\right)=\sqrt{3} + i \\
for\,k=1\\
\sqrt[2]{4}\left(\cos\left(\frac{\left(\frac{\pi}{3}\right)+2\pi \cdot 1}{2}\right)+i\sin\left(\frac{\left(\frac{\pi}{3}\right)+2\pi \cdot 1}{2}\right)\right)\\=2\left(\cos\left(\frac{7 \pi}{6}\right)+i\sin\left(\frac{7 \pi}{6}\right)\right)=- \sqrt{3} - i \\
$
$so\,\,\\
\sqrt[2]{2 + 2 \sqrt{3} i}=\sqrt{3}+i \\
\sqrt[2]{2 + 2 \sqrt{3} i}=-\sqrt{3}-i
$
