Key Concepts and Formulas

• Roots of a Quadratic Equation: For the quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$.
• Complex Number Properties: $i = \sqrt{-1}$, $i^2 = -1$. $\operatorname{Re}(a + bi) = a$ and $\operatorname{Im}(a + bi) = b$.
• Binomial Theorem: $(a \pm b)^2 = a^2 \pm 2ab + b^2$

Step-by-Step Solution

1. Manipulating the Given Quadratic Equation

We are given the quadratic equation $2z^2 - 3z - 2i = 0$. Let $\alpha$ and $\beta$ be its roots. Our goal is to find a useful relationship involving $\alpha$ and $\beta$. Since $z=0$ is not a root, we can divide the equation by $z$:

$2z - 3 - \frac{2i}{z} = 0$

Rearranging the terms, we get:

$2z - \frac{2i}{z} = 3$

$2\left(z - \frac{i}{z}\right) = 3$

Dividing by 2, we obtain a crucial relation:

$z - \frac{i}{z} = \frac{3}{2} \quad (*)$

This relation holds for both roots $\alpha$ and $\beta$:

$\alpha - \frac{i}{\alpha} = \frac{3}{2}$

$\beta - \frac{i}{\beta} = \frac{3}{2}$

Why this step? Expressing the equation in terms of $z - i/z$ creates a simpler building block to generate higher powers. This symmetric form is often useful in problems involving powers of roots.

2. Deriving a Relation for $z^2 - 1/z^2$

Now, we square the relation $(*)$ to find an expression involving $z^2$:

$\left( z - \frac{i}{z} \right)^2 = \left( \frac{3}{2} \right)^2$

Expanding the left side using $(a-b)^2 = a^2 + b^2 - 2ab$:

$z^2 + \left(\frac{i}{z}\right)^2 - 2 \cdot z \cdot \left(\frac{i}{z}\right) = \frac{9}{4}$

Simplifying the terms: $(i/z)^2 = i^2/z^2 = -1/z^2$, and $2 \cdot z \cdot (i/z) = 2i$.

$z^2 - \frac{1}{z^2} - 2i = \frac{9}{4}$

Rearranging to isolate $z^2 - 1/z^2$:

$z^2 - \frac{1}{z^2} = \frac{9}{4} + 2i$

This relation also holds for both $\alpha$ and $\beta$.

Why this step? We are progressively generating higher power symmetric expressions ($z^n \pm 1/z^n$) which are likely to appear in the target expression.

3. Deriving a Relation for $z^4 + 1/z^4$

We now square the expression for $z^2 - 1/z^2$:

$\left( z^2 - \frac{1}{z^2} \right)^2 = \left( \frac{9}{4} + 2i \right)^2$

Expanding the left side using $(a-b)^2 = a^2 + b^2 - 2ab$:

$(z^2)^2 + \left(\frac{1}{z^2}\right)^2 - 2 \cdot z^2 \cdot \left(\frac{1}{z^2}\right) = z^4 + \frac{1}{z^4} - 2$

Expanding the right side using $(a+b)^2 = a^2 + b^2 + 2ab$:

$\left(\frac{9}{4}\right)^2 + (2i)^2 + 2 \cdot \frac{9}{4} \cdot (2i) = \frac{81}{16} - 4 + 9i$

Equating both sides:

$z^4 + \frac{1}{z^4} - 2 = \frac{81}{16} - 4 + 9i$

Rearranging to isolate $z^4 + 1/z^4$:

$z^4 + \frac{1}{z^4} = \frac{81}{16} - 4 + 2 + 9i$

$z^4 + \frac{1}{z^4} = \frac{81}{16} - 2 + 9i$

To combine the real parts, find a common denominator:

$z^4 + \frac{1}{z^4} = \frac{81}{16} - \frac{32}{16} + 9i$

$z^4 + \frac{1}{z^4} = \frac{49}{16} + 9i$

This identity is very important, as it holds for both $\alpha$ and $\beta$:

$\alpha^4 + \frac{1}{\alpha^4} = \frac{49}{16} + 9i$

$\beta^4 + \frac{1}{\beta^4} = \frac{49}{16} + 9i$

Let's denote this common value as $K = \frac{49}{16} + 9i$.

Why this step? The powers in the given expression ($\alpha^{19}$, $\alpha^{11}$) differ by 8, suggesting that $\alpha^8$ or related terms might be a factor. By computing $z^4 + 1/z^4$, we anticipate using this in the simplification.

4. Simplifying the Given Complex Expression

The expression we need to evaluate is:

$E = \frac{\alpha^{19}+\beta^{19}+\alpha^{11}+\beta^{11}}{\alpha^{15}+\beta^{15}}$

We can factor terms in the numerator. Observe the powers: $19$ and $11$. Their difference is $8$. If we factor out $\alpha^{15}$, we get:

$\alpha^{19} + \alpha^{11} = \alpha^{15} \cdot \alpha^4 + \alpha^{15} \cdot \alpha^{-4} = \alpha^{15} \left( \alpha^4 + \frac{1}{\alpha^4} \right)$

Similarly for $\beta$:

$\beta^{19} + \beta^{11} = \beta^{15} \cdot \beta^4 + \beta^{15} \cdot \beta^{-4} = \beta^{15} \left( \beta^4 + \frac{1}{\beta^4} \right)$

Substitute these back into the expression for $E$:

$E = \frac{\alpha^{15} \left( \alpha^4 + \frac{1}{\alpha^4} \right) + \beta^{15} \left( \beta^4 + \frac{1}{\beta^4} \right)}{\alpha^{15}+\beta^{15}}$

From Step 3, we know that $\alpha^4 + 1/\alpha^4 = K$ and $\beta^4 + 1/\beta^4 = K$. Substitute $K$ into the expression:

$E = \frac{\alpha^{15} K + \beta^{15} K}{\alpha^{15}+\beta^{15}}$

Factor out $K$ from the numerator:

$E = \frac{K (\alpha^{15} + \beta^{15})}{\alpha^{15}+\beta^{15}}$

Assuming $\alpha^{15}+\beta^{15} \neq 0$ (which is a safe assumption in such problems unless specified), we can cancel the common term:

$E = K$

Therefore, the complex expression simplifies to:

$E = \frac{49}{16} + 9i$

Why this step? Recognizing common factors and the symmetric nature of the derived expression $z^4 + 1/z^4$ is key to simplifying the intimidating-looking fraction. This avoids the need to calculate the actual values of $\alpha^{15}$ or $\beta^{15}$.

5. Calculating the Final Required Value

We need to calculate $16 \cdot \operatorname{Re}(E) \cdot \operatorname{Im}(E)$.

From $E = \frac{49}{16} + 9i$:

The real part is $\operatorname{Re}(E) = \frac{49}{16}$.

The imaginary part is $\operatorname{Im}(E) = 9$.

Now, substitute these values into the final expression:

$16 \cdot \operatorname{Re}(E) \cdot \operatorname{Im}(E) = 16 \cdot \left( \frac{49}{16} \right) \cdot 9$

The $16$ in the numerator and denominator cancel out:

$= 49 \cdot 9$

$= 441$

Common Mistakes & Tips

• Forgetting to square both sides of the equation when manipulating expressions.
• Not recognizing the symmetry in the expressions and failing to factor out common terms.
• Making arithmetic errors while simplifying complex numbers.

Summary

By cleverly manipulating the given quadratic equation and recognizing the symmetry in the derived expressions, we simplified a complex expression involving powers of the roots. We found that the expression simplifies to $\frac{49}{16} + 9i$, and then calculated $16 \cdot \operatorname{Re}(E) \cdot \operatorname{Im}(E) = 441$.

Final Answer

The final answer is $\boxed{441}$, which corresponds to option (A).