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.
On
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.)
On
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 $-(2-\sqrt 3\mathrm i)$: $$u+\mathrm iv=\pm \mathrm i(2-\sqrt 3\mathrm i)=\color{red}{\pm\dfrac{\sqrt2}2\bigl(\sqrt{-2+\sqrt 7}+\mathrm i\sqrt{2+\sqrt 7}\bigr)}.$$
On
$$\text{z}=\sqrt{\sqrt{1-4i\sqrt{3}}}=\left(\sqrt{1-4i\sqrt{3}}\right)^{\frac{1}{2}}=\left(\left(1-4i\sqrt{3}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=$$ $$\left(1-4i\sqrt{3}\right)^{\frac{1}{4}}=\left(7e^{-\arctan\left(4\sqrt{3}\right)i}\right)^{\frac{1}{4}}=\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}$$
So:
$$\Re\left[\text{z}\right]=\Re\left[\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}\right]=\sqrt[4]{7}\cos\left(-\frac{\arctan\left(4\sqrt{3}\right)}{4}\right)$$ $$\Im\left[\text{z}\right]=\Im\left[\sqrt[4]{7}e^{-\frac{\arctan\left(4\sqrt{3}\right)i}{4}}\right]=\sqrt[4]{7}\sin\left(-\frac{\arctan\left(4\sqrt{3}\right)}{4}\right)$$
On
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.
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.