Key Concepts and Formulas

• Complex Numbers: A complex number is of the form $z = a + bi$, where $a$ and $b$ are real numbers, and $i = \sqrt{-1}$.
• Square Root of a Complex Number: If $z = a + bi$, then $\sqrt{z}$ is a complex number $x + yi$ such that $(x + yi)^2 = a + bi$. This leads to the equations $x^2 - y^2 = a$ and $2xy = b$.
• Complex Conjugate: The complex conjugate of $z = a + bi$ is $z^* = a - bi$. If $w$ is a square root of $z$, then $w^*$ is a square root of $z^*$.

Step-by-Step Solution

Step 1: Simplify the terms inside the square roots

The expression is: ${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$ Simplify $\sqrt{-54}$: $\sqrt{-54} = \sqrt{54} \cdot \sqrt{-1} = \sqrt{9 \cdot 6} \cdot i = 3\sqrt{6}i$ Substitute this back into the original expression: ${\left( {3 + 2(3\sqrt{6}i)} \right)^{{1 \over 2}}} - {\left( {3 - 2(3\sqrt{6}i)} \right)^{{1 \over 2}}}$ $= {\left( {3 + 6\sqrt{6}i} \right)^{{1 \over 2}}} - {\left( {3 - 6\sqrt{6}i} \right)^{{1 \over 2}}}$ This step simplifies the complex numbers within the square roots, making them easier to manipulate.

Step 2: Find the square roots of $3 + 6\sqrt{6}i$

We want to find a complex number $x + yi$ such that $(x + yi)^2 = 3 + 6\sqrt{6}i$. Expanding the square gives: $x^2 + 2xyi - y^2 = 3 + 6\sqrt{6}i$ Equating the real and imaginary parts, we get: $x^2 - y^2 = 3$ $2xy = 6\sqrt{6}$ $xy = 3\sqrt{6}$ We look for integer/simple solutions. Consider $x=3$ and $y = \sqrt{6}$. Then $x^2 - y^2 = 9 - 6 = 3$, which satisfies the real part equation. Thus, $3 + \sqrt{6}i$ is a square root of $3 + 6\sqrt{6}i$. Therefore, the square roots of $3 + 6\sqrt{6}i$ are $\pm(3 + \sqrt{6}i)$. ${\left( {3 + 6\sqrt{6}i} \right)^{{1 \over 2}}} = \pm \left( {3 + \sqrt{6}i} \right)$ Expressing the complex number as a perfect square simplifies the square root calculation.

Step 3: Find the square roots of $3 - 6\sqrt{6}i$

Notice that $3 - 6\sqrt{6}i$ is the complex conjugate of $3 + 6\sqrt{6}i$. Since the square roots of $3 + 6\sqrt{6}i$ are $\pm(3 + \sqrt{6}i)$, the square roots of $3 - 6\sqrt{6}i$ are $\pm(3 - \sqrt{6}i)$. ${\left( {3 - 6\sqrt{6}i} \right)^{{1 \over 2}}} = \pm \left( {3 - \sqrt{6}i} \right)$ Using the property of complex conjugates avoids redundant calculations.

Step 4: Evaluate the expression considering the relationship between the square roots

Let $A = {\left( {3 + 6\sqrt{6}i} \right)^{{1 \over 2}}}$ and $B = {\left( {3 - 6\sqrt{6}i} \right)^{{1 \over 2}}}$. Then $A \in \{ (3 + \sqrt{6}i), -(3 + \sqrt{6}i) \}$ and $B \in \{ (3 - \sqrt{6}i), -(3 - \sqrt{6}i) \}$. We need to find the possible values of $A - B$. Consider $A^2B^2 = (3+6\sqrt{6}i)(3-6\sqrt{6}i) = 9 + 36(6) = 9 + 216 = 225$. Thus $AB = \pm 15$.

Let $w_1 = 3 + \sqrt{6}i$ and $w_2 = 3 - \sqrt{6}i$. Then $w_1 w_2 = (3 + \sqrt{6}i)(3 - \sqrt{6}i) = 9 + 6 = 15$.

Case 1: $AB = 15$

• If $A = w_1 = (3 + \sqrt{6}i)$, then $B = w_2 = (3 - \sqrt{6}i)$. Then $A - B = (3 + \sqrt{6}i) - (3 - \sqrt{6}i) = 2\sqrt{6}i$.
• If $A = -w_1 = -(3 + \sqrt{6}i)$, then $B = -w_2 = -(3 - \sqrt{6}i)$. Then $A - B = -(3 + \sqrt{6}i) - (-(3 - \sqrt{6}i)) = -2\sqrt{6}i$.

Case 2: $AB = -15$

• If $A = w_1 = (3 + \sqrt{6}i)$, then $B = -w_2 = -(3 - \sqrt{6}i)$. Then $A - B = (3 + \sqrt{6}i) - (-(3 - \sqrt{6}i)) = 6$.
• If $A = -w_1 = -(3 + \sqrt{6}i)$, then $B = w_2 = (3 - \sqrt{6}i)$. Then $A - B = -(3 + \sqrt{6}i) - (3 - \sqrt{6}i) = -6$.

Therefore, the possible values of the expression are $2\sqrt{6}i, -2\sqrt{6}i, 6, -6$.

The constraint on $AB$ is crucial for finding the correct values.

Step 5: Identify the possible imaginary parts and compare with options

The possible values of the expression are $2\sqrt{6}i$, $-2\sqrt{6}i$, $6$, and $-6$. The imaginary parts are $2\sqrt{6}$, $-2\sqrt{6}$, $0$, and $0$.

Comparing with the options: (A) $-2\sqrt{6}$ (B) $6$ (C) $\sqrt{6}$ (D) $-\sqrt{6}$

Option (A) is one of the possible imaginary parts.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when simplifying complex numbers and finding square roots.
• Independent Square Roots: Avoid treating the square roots as completely independent. The product of the terms inside the square roots provides a constraint.
• Complex Conjugates: Utilizing complex conjugates can save time and effort.

Summary

We simplified the complex numbers, found their square roots using algebraic manipulation and the conjugate property, and then considered the constraint on the product of the square roots to determine the possible values of the expression. The imaginary part of the expression can be $-2\sqrt{6}$.

The final answer is \boxed{-2\sqrt{6}}, which corresponds to option (A).