# 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

$$T = 3 + 2\sqrt { - 54} = 3 + i6\sqrt { 6} \to {I^{st}} - Quadrant - Angle = \theta$$

$$U = 3 - 2\sqrt { - 54} = 3 - i6\sqrt { 6} \to I{V^{th}} - Quadrant - Angle = - \theta$$

$$\Rightarrow \left( {{{\left( {3 + i6\sqrt 6 } \right)}^{\frac{1}{2}}} - {{\left( {3 - i6\sqrt 6 } \right)}^{\frac{1}{2}}}} \right)$$

$$r\cos \theta = 3$$ & $$r\sin \theta = 6\sqrt 6 \Rightarrow {r^2} = 225 \Rightarrow r = 15 \Rightarrow \tan \theta = 2\sqrt 6$$

$$\Rightarrow \left( {\sqrt {15} {e^{\frac{{i\theta }}{2}}} - \sqrt {15} {e^{ - \frac{{i\theta }}{2}}}} \right) \Rightarrow \sqrt {15} \left( {{e^{\frac{{i\theta }}{2}}} - {e^{ - \frac{{i\theta }}{2}}}} \right) = i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right)$$

$$\Rightarrow \frac{{2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\frac{\theta }{2}}} = 2\sqrt 6 \Rightarrow {\tan ^2}\frac{\theta }{2} + \frac{2}{{\sqrt {24} }}\tan \frac{\theta }{2} + \frac{1}{{24}} = \frac{{25}}{{24}} \Rightarrow \left( {\tan \frac{\theta }{2} + \frac{1}{{\sqrt {24} }}} \right) = \frac{5}{{\sqrt {24} }} \Rightarrow \tan \frac{\theta }{2} = \frac{4}{{\sqrt {24} }} = \frac{{\sqrt 2 }}{{\sqrt 3 }}$$

$$\sin \frac{\theta }{2} = \frac{{\sqrt 3 }}{{\sqrt 5 }} \Rightarrow i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right) = i\sqrt {15} \left( {2 \times \frac{{\sqrt 2 }}{{\sqrt 5 }}} \right) = 2\sqrt 6 i$$

• This question appeared in the entrance examination and I cross checked the question twice Sep 4, 2020 at 10:24
• How do you define $z^{1/2}$? Sep 4, 2020 at 10:26
• $z = r{e^{i\theta }}\because {z^{\frac{1}{2}}} = {r^{\frac{1}{2}}}{e^{\frac{{i\theta }}{2}}}$ Sep 4, 2020 at 10:32
• @SamarImamZaidi How do you define (uniquely) $\theta$ in $z=re^{i\theta}$ ? Sep 4, 2020 at 10:35
• $z = x + iy \Rightarrow \theta = {\tan ^{ - 1}}\frac{y}{x}$ Sep 4, 2020 at 10:47

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.

• There is rather a sweet generalization. You could mention it here :) Sep 4, 2020 at 11:12

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 $$x+iy=\pm(3-i\sqrt 6).$$

• "Le bon vieux truc !". Cheers :-) Sep 4, 2020 at 11:36
• @ClaudeLeibovici: I remember it was in a maths book for the class of ‘philo’ some decades ago. Sep 4, 2020 at 11:40