Finding the difference of square root of conjugate complex number Find the imaginary part of $\left( {{{\left( {3 + 2\sqrt { - 54} } \right)}^{\frac{1}{2}}} - {{\left( {3 - 2\sqrt { - 54} } \right)}^{\frac{1}{2}}}} \right)$
(1) $-\sqrt 6$
(2) $-2\sqrt 6$
(3) $\sqrt 6$
(4) $6$
My Approach is as follow and none of the answer is matching, I cross checked it
$T = 3 + 2\sqrt { - 54}  = 3 + i6\sqrt {  6}  \to {I^{st}} - Quadrant - Angle = \theta $
$U = 3 - 2\sqrt { - 54}  = 3 - i6\sqrt {  6}  \to I{V^{th}} - Quadrant - Angle =  - \theta $
$\Rightarrow \left( {{{\left( {3 + i6\sqrt 6 } \right)}^{\frac{1}{2}}} - {{\left( {3 - i6\sqrt 6 } \right)}^{\frac{1}{2}}}} \right)$
$r\cos \theta  = 3$ & $r\sin \theta  = 6\sqrt 6  \Rightarrow {r^2} = 225 \Rightarrow r = 15 \Rightarrow \tan \theta  = 2\sqrt 6 $
$ \Rightarrow \left( {\sqrt {15} {e^{\frac{{i\theta }}{2}}} - \sqrt {15} {e^{ - \frac{{i\theta }}{2}}}} \right) \Rightarrow \sqrt {15} \left( {{e^{\frac{{i\theta }}{2}}} - {e^{ - \frac{{i\theta }}{2}}}} \right) = i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right)$
$ \Rightarrow \frac{{2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\frac{\theta }{2}}} = 2\sqrt 6  \Rightarrow {\tan ^2}\frac{\theta }{2} + \frac{2}{{\sqrt {24} }}\tan \frac{\theta }{2} + \frac{1}{{24}} = \frac{{25}}{{24}} \Rightarrow \left( {\tan \frac{\theta }{2} + \frac{1}{{\sqrt {24} }}} \right) = \frac{5}{{\sqrt {24} }} \Rightarrow \tan \frac{\theta }{2} = \frac{4}{{\sqrt {24} }} = \frac{{\sqrt 2 }}{{\sqrt 3 }}$
$\sin \frac{\theta }{2} = \frac{{\sqrt 3 }}{{\sqrt 5 }} \Rightarrow i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right) = i\sqrt {15} \left( {2 \times \frac{{\sqrt 2 }}{{\sqrt 5 }}} \right) = 2\sqrt 6 i$

It is simpler to calculate the square roots of a complex number by hand: denote such a square root as $z=x+iy$. Then $z^2=x^2-y^2+2ixy$, so for the square roots of $3+2\sqrt{-54}=3+6i\sqrt 6$, you have to solve the system of equations
\begin{cases}
x^2-y^2=3, \\
xy=3\sqrt 6.
\end{cases}
Now there's a trick to make the computation faster:
$$|z|^2=x^2+y^2=\bigl|3+6i\sqrt 6\bigr|=\sqrt{225}=15,$$
so we have a linear system in $x^2$ and $y^2$: $\;\smash[b]{\begin{cases}
x^2-y^2=3, \\ x^2+y^2 =15,\end{cases}}$ and the  3rd equation tells us that $x$ and $y$ have the same sign.
Solving this linear system, we obtain
$$x^2=9,\enspace y^2=6,\quad\text{whence}\quad x+iy=\pm(3+i\sqrt 6).$$
For the square roots of $3-2\sqrt{-54}=3-6i\sqrt 6$, we obtain the same linear system, with the condition that $x$ and $y$ have opposite signs, so that the square roots are
$$x+iy=\pm(3-i\sqrt 6).$$
