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.

You need to first find the square root of $1 - 4 \sqrt{3}i$ using the same method: let $(c+di)^2 = 1 - 4 \sqrt{3} i$ and then compare real and imaginary parts to find $c$ and $d$ explicitly. This will give you two different answers. Then assume $(a+bi)^2 = c + di$, and compare real and imaginary parts to find $a$ and $b$ explicitly. This gives you two solutions for both solutions of $(c+di)^2 = 1 - 4 \sqrt{3} i$, so in total, you get four different answers.
