(a+bi)^2=4+4sqrt3i
I got
a^2-b^2+2abi=4+4sqrt3i
a^2=b^2+4
a=sqrt(b^2+4)
2ab=4sqrt3
a=4sqrt3/2b
I can't seem to get the answer when I compare the values
The answer given for a is +/- sqrt 6 and b is +/- sqrt2
$\displaystyle a= 2\sqrt{3}/b$
I would avoid square roots and start from this. Now $\displaystyle a^2= 12/b^2$ so that $\displaystyle a^2+ b^2= 4$ becomes $\displaystyle \frac{12}{b^2}+ b^2= 4$. Multiply both sides by $\displaystyle b^2$ to get $\displaystyle 12+ b^4= 4b^2$ of $\displaystyle (b^2)^2- 4b^2+ 12= 0$.
That is a quadratic equation of $\displaystyle b^2$- and easily factorable. You will find two values for $\displaystyle b^2$ so 4 different values for b.
I can't seem to get the answer when I compare the values
The answer given for a is +/- sqrt 6 and b is +/- sqrt2
This might be easiest to solve using polars.
$\displaystyle \displaystyle (a + ib)^2 = 4 + 4\sqrt{3}i = 8e^{i\frac{\pi}{3}}$, so
$\displaystyle \displaystyle \begin{align*}a + ib &= \pm\left(8e^{i\frac{\pi}{3}}\right)^{\frac{1}{2}} \\ &= \pm 2\sqrt{2}e^{i\frac{\pi}{6}}\\ &= \pm 2\sqrt{2}\left(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}\right) \\ &= \pm 2\sqrt{2}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) \\ &= \pm\left(\sqrt{6} + i\sqrt{2}\right) \end{align*}$