Nice answer proyaop, but squaring either one of those doesn't end up with 2i+2

\(2+2i\\ =2\sqrt{2}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ =2\sqrt{2}(\cos(45^\circ)+i\sin(45^\circ))\)

when taking the square root, the argument divides by 2 and the magnitude square roots, so it is:

\(=2^{\frac{3}{4}}(\cos(22.5^\circ)+i\sin(22.5^\circ))\)

To solve for \(\cos(22.5^\circ)\) and \(\sin(22.5^\circ)\), we can make use of the half-angle identity:

\(\cos(22.5^\circ)\\= \sqrt{\frac{1+\cos(45^\circ)}{2}}\\= \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2+\sqrt{2}}{4}}\\= \frac{\sqrt{2+\sqrt{2}}}{2}\)

(it's positive because it's in the first quadrant)

Same thing with sin:

\(\sin(22.5^\circ)\\= \sqrt{\frac{1-\cos(45^\circ)}{2}}\\= \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\= \sqrt{\frac{2-\sqrt{2}}{4}}\\= \frac{\sqrt{2-\sqrt{2}}}{2}\)

Thus, the final root is:

\(2^{\frac{3}{4}}(\frac{\sqrt{2+\sqrt{2}}}{2}+\frac{\sqrt{2-\sqrt{2}}}{2}i)\\ =\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}+\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i \)

The simplifying is up to you.

Also, there is one more root, if you use \(\cos(22.5^\circ+180^\circ)\) and \(\sin(22.5^\circ+180^\circ)i\), because there are 2 roots of unity. Notice that the cosine and sine flip signs in that case, because they get to the 3rd quadrant. So the other root is \(=-\frac{\sqrt{2+\sqrt{2}}}{\sqrt[4]{2}}-\frac{\sqrt{2-\sqrt{2}}}{\sqrt[4]{2}}i\)

I seriously doubt that the problem was meant to be this complicated