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 step:
Let $x+\mathrm i y$ a square root of $1-4\sqrt 3\mathrm i$ This means $\;x^2-y^2+2xy\mathrm i=1-4\sqrt 3\mathrm i$, whence
$$x^2-y^2=1,\quad xy=-2\sqrt 3.$$
Furthermore, $\lvert x+\mathrm i y\rvert^2=x^2+y^2=\lvert 1-4\sqrt 3\mathrm i \rvert=\sqrt{49}$.
So we have to solve the linear system (in $x^2$ and $y^2$):
$$\begin{cases}x^2-y^2=1,\\x^2+y^2=7\end{cases}\iff x^2= 4,\enspace y^2=3.$$
Thus, $x=\pm 2$, $y=\pm\sqrt3$. Finally, $xy$ is negative ($-2\sqrt3$), which implies the solutions are
$$x+\mathrm i y=\pm(2-\sqrt 3\mathrm i).$$
Second step:
Let $u+\mathrm iv$  a square root of $2-\sqrt 3\mathrm i$.  The same method leads to the system 
$$\begin{cases}u^2-v^2=2,\\2uv=-\sqrt 3,\\u^2+v^2=\sqrt 7.\end{cases}$$
We deduce $u^2=\dfrac{2+\sqrt 7}2$, $\;v^2=\dfrac{-2+\sqrt 7}2$, $\;uv<0$, so that the first two roots:
$$u+\mathrm iv=\color{red}{\pm\dfrac{\sqrt2}2\bigl(\sqrt{2+\sqrt 7}-\mathrm i\sqrt{-2+\sqrt 7}\bigr)}.$$
The other two roots correspond to $-(2-\sqrt 3\mathrm i)$:
$$u+\mathrm iv=\pm \mathrm i(2-\sqrt 3\mathrm i)=\color{red}{\pm\dfrac{\sqrt2}2\bigl(\sqrt{-2+\sqrt 7}+\mathrm i\sqrt{2+\sqrt 7}\bigr)}.$$
