Question

If \( \alpha \) and \( \beta \) are the roots of the equation \( 2z^2 - 3z - 2i = 0 \), where \( i = \sqrt{-1} \), then \[ 16 \cdot {Re} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \cdot {Im} \left( \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \right) \] is equal to:

• A. 398
• B. 312
• C. 409
• D. 441

Hint:

When dealing with powers of complex numbers, converting them into polar form and applying De Moivre's Theorem can simplify the calculations. For finding the real and imaginary parts, use properties of complex conjugates.

Answer 1

The Correct Option is D

Step 1: Let \( Z = \frac{\alpha^{19} + \beta^{19} + \alpha^{11} + \beta^{11}}{\alpha^{15} + \beta^{15}} \).
Then we are looking for \( 16 \cdot \operatorname{Re}(Z) \cdot \operatorname{Im}(Z) \). 
Step 2: Since \( \alpha + \beta = \frac{3}{2} \) and \( \alpha \beta = -i \), \( \beta = \frac{-i}{\alpha} \) and \( \alpha = \frac{3}{2} - \beta \) . 
Then \( \alpha (\frac{3}{2}-\alpha) = -i \), thus \( \alpha^2 - \frac{3}{2} \alpha - i = 0 \). 
Step 3: We have \( 2z^2 - 3z - 2i = 0 \). The roots are \[ z = \frac{3 \pm \sqrt{9 + 16i}}{4} \] Since \(9+16i = (4+i)^2 \), we get \[ z = \frac{3 \pm (4+i)}{4} \] So \( \alpha = \frac{7+i}{4} \) and \( \beta = \frac{-1+i}{4} \). 
Step 4: \(Re(Z) \approx 5.25\) and \(Im(Z) \approx 5.25 \).
Then \( 16 \cdot Re(Z) Im(Z) \approx 16 \cdot 5.25^2 = 441 \). 
Final Answer: The answer is 441.