Animated Solution for Mathematics - Complex Numbers: The imaginary part of $(3+2\sqrt{-54}{)}^{1/2}-(3-2\sqrt{-54}{)}^{1/2}$ can be

Imagine you are standing at the edge of a mathematical cliff. You are looking at the expression $(3+2\sqrt{-54}{)}^{\frac{1}{2}}-(3-2\sqrt{-54}{)}^{\frac{1}{2}}$.

The negative sign inside the square root is a classic signpost in JEE Advanced problems, signaling that we are leaving the comfortable world of real numbers and entering the elegant, multidimensional realm of complex numbers. Let's take a deep breath and break this down.

Decoding the Complex Number

First, let's tackle that $\sqrt{-54}$. We know that $i=\sqrt{-1}$.

So, $\sqrt{-54}=\sqrt{54}\times i$. Since $54=9\times 6$, we have $\sqrt{54}=3\sqrt{6}$.

Thus, $\sqrt{-54}=3i\sqrt{6}$. Our expression now becomes:

This is much cleaner. We have two complex numbers, $z_1=3+6i\sqrt{6}$ and $z_2=3-6i\sqrt{6}$.

Notice that $z_2$ is simply the complex conjugate of $z_1$. This symmetry is our secret weapon.

The Square Root Algorithm

To find the square root of $z_1=3+6i\sqrt{6}$, we assume $\sqrt{3+6i\sqrt{6}}=a+ib$, where $a$ and $b$ are real numbers. Squaring both sides gives us:

By equating the real and imaginary parts, we get two fundamental equations:

1. $a^2-b^2=3$

2. $2ab=6\sqrt{6}\Rightarrow ab=3\sqrt{6}$

The Modulus Trick

Now, we need a third equation to solve for $a$ and $b$ without getting lost in substitution. We use the modulus property: $|a+ib{|}^2=|3+6i\sqrt{6}|$.

This gives us:

Now we have a system of two equations: $a^2-b^2=3$ and $a^2+b^2=15$.

Adding them gives $2a^2=18$, so $a^2=9$, meaning $a=\pm 3$. Subtracting them gives $2b^2=12$, so $b^2=6$, meaning $b=\pm \sqrt{6}$.

Since $ab=3\sqrt{6}>0$, $a$ and $b$ must have the same sign. Thus, the square roots of $z_1$ are $\pm (3+i\sqrt{6})$.

The Final Evaluation

Since $z_2$ is the conjugate of $z_1$, its square roots are simply the conjugates of the roots of $z_1$: $\pm (3-i\sqrt{6})$.

Now, we evaluate the original expression: $\pm (3+i\sqrt{6})-\pm (3-i\sqrt{6})$.

If we choose the positive root for both, we get:

If we choose the negative root for both, we get:

The imaginary part is either $2\sqrt{6}$ or $-2\sqrt{6}$. Looking at our options, $-2\sqrt{6}$ is the clear winner. You have navigated the complex plane and emerged victorious!