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.
The answer depends in part on the meaning that's ascribed to the square root symbol, $\sqrt{}$.  In almost all settings, $\sqrt a$ is taken to be positive if $a$ is a positive real number.  Assuming this to be the case, then the square root of the square root of $1-4\sqrt3i$ turns out to be one of the four number $z$, $zi$, $-z$, or $-zi$, where

$$z=\sqrt{{\sqrt7+2\over2}}-\sqrt{{\sqrt7-2\over2}}i$$

Exactly which one is meant depends on the convention you adopt for assigning an unambiguous value to $\sqrt{a+bi}$ if $a,b\in\mathbb{R}$ and $b\not=0$.  (Most conventions set $\sqrt a=\sqrt{-a}i$ if $a$ is a negative real number. Note that $\sqrt7\gt2$, so all values inside square root symbols in the displayed equation are positive.)  Let me explain what I mean by this, and in the process show how the expression for $z$ was obtained.

One convention says that $\sqrt{}$ applied to a complex number with non-zero imaginary part has a positive real part. (This corresponds to assuming $-\pi\lt\theta\le\pi$ in the equation $\sqrt{e^{i\theta}}=e^{i\theta/2}.)$  Under that convention we have

$$\sqrt{1-4\sqrt3i}=2-\sqrt3i$$

so it remains to find real values $a$ and $b$ with $a\gt0$ such that $(a+bi)^2=2-\sqrt3i$, which is to say

$$a^2-b^2=2\qquad\text{and}\qquad2ab=-\sqrt3$$

Plugging $b=-\sqrt3/(2a)$ into $a^2-b^2=2$ leads to

$$4a^4-8a^2-3=0$$

which implies $a^2=(2\pm\sqrt7)/2$.  In order for $a$ to be real, we must have $a^2=(2+\sqrt7)/2$, hence

$$a=\sqrt{{\sqrt7+2\over2}}$$

The corresponding value for $b$ is most easily obtained by noting that $a^2-b^2=2$ implies $|b|=\sqrt{a^2-2}$ while $b=-\sqrt3/(2a)$ implies $b$ is negative, hence

$$b=-\sqrt{a^2-2}=-\sqrt{{\sqrt7+2\over2}-2}=-\sqrt{{\sqrt7-2\over2}}$$

which gives us the answer

$$\sqrt{\sqrt{1-4\sqrt3i}}=\sqrt{{\sqrt7+2\over2}}-\sqrt{{\sqrt7-2\over2}}i$$

Another convention says that $\sqrt{}$ applied to a complex number with non-zero imaginary part has a positive imaginary part. (This corresponds to assuming $0\le\theta\lt2\pi$ in the equation $\sqrt{e^{i\theta}}=e^{i\theta/2}$.)  Under this convention we have

$$\sqrt{1-4\sqrt3i}=-2+\sqrt3i$$

and, if you carry out the same reasoning as before, this time with  $a^2-b^2=-2$ and $2ab=\sqrt3$, you arrive at

$$\sqrt{\sqrt{1-4\sqrt3i}}=\sqrt{{\sqrt7-2\over2}}+\sqrt{{\sqrt7+2\over2}}i$$

which can be recognized as $zi$.  (We can check that we have the correct signs by noting that $2ab=\sqrt3$ implies $a$ must have the same sign as $b$, which the convention requires to be positive.)

A final remark:  My thanks to Wojowu, who pointed out an egregious mistake in my original answer.  (I thought the number in question was $\sqrt{1-4\sqrt3}i$ rather than $\sqrt{1-4\sqrt3i}$, which, if you can see the difference, makes a world of difference.)