Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number 
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part? 
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-4\sqrt{3}i}=\sqrt{2^2+3i^2-4\sqrt{3}i}=\sqrt{(2-\sqrt{3}i)^2}$ which made calculation easier.

First establish a formula for the square root:
$$(a+ib)^2=a^2-b^2+i2ab=x+iy$$ is equivalent to the system
$$a^2-b^2=x,\\2ab=y.$$
If we notice that $$(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2,$$ we have
$$a^2-b^2=x,\\
a^2+b^2=\pm\sqrt{x^2+y^2}.$$
The real solutions are
$$a=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}2},\\b=\pm\sqrt{\frac{\sqrt{x^2+y^2}-x}2},$$
where the signs must be chosen so that $\text{sign}(a)\text{sign}(b)=\text{sign}(y)$.

Applying to the given case, $x=1,y=-4\sqrt3$, we find $a=\pm2,b=\mp\sqrt3$.
Then once again with $x=\pm2,y=\mp\sqrt3$, we get the final answer
$$a=\pm\sqrt{\frac{\sqrt{7}+2}2},b=\mp\sqrt{\frac{\sqrt{7}-2}2},$$ or
$$a=\pm\sqrt{\frac{\sqrt{7}-2}2},b=\pm\sqrt{\frac{\sqrt{7}+2}2}.$$
These four solutions form a square.
