<s>[INST] 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 }}$
$\sin \frac{\theta }{2} = \frac{{\sqrt 3 }}{{\sqrt 5 }} \Rightarrow i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right) = i\sqrt {15} \left( {2 \times \frac{{\sqrt 2 }}{{\sqrt 5 }}} \right) = 2\sqrt 6 i$
 [/INST] Let $$z=\sqrt{3+6\sqrt{6}i)}=x+iy ~~~(1)$$
Squaring we get $3+6\sqrt{6}i=x^2-y^2+2ixy$
$$\implies x^2-y^2=3, xy=3\sqrt{6}~~~~(2)$$
$$\bar z=\sqrt{3-6\sqrt{6}i}=x-iy~~~~(3),$$
multiplying the two (1) and (3)we get $$x^2+y^2=\sqrt{9+36.6}=15~~~(4)$$
Using (4) in (2) we get $x=\pm 3, y=\pm \sqrt{6}.$
So $z-\bar z=2y=\pm 2\sqrt{6}.$
Hence, option (B) is one correct answer.
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