Question

If \( Z = \frac{2 - i}{\alpha + i} \) and \( 4 \, \text{Re}(Z) = 3 \, \text{Im}( \overline{Z} ) \), find \( \alpha \).

• A. \( \alpha = -2 \)
• B. \( \alpha = 3 \)
• C. \( \alpha = -3 \)
• D. \( \alpha = 2 \)

Hint:

When working with complex numbers, use conjugates to simplify and separate real and imaginary parts for easier calculation.

Answer 1

The Correct Option is A

We are given the complex number \( Z = \frac{2 - i}{\alpha + i} \), and the condition that \( 4 \, \text{Re}(Z) = 3 \, \text{Im}( \overline{Z} ) \).

1. Step 1: Express \( Z \) in terms of real and imaginary parts. To simplify the expression for \( Z \), multiply both the numerator and denominator of \( Z \) by the conjugate of the denominator \( \alpha - i \): \[ Z = \frac{2 - i}{\alpha + i} \cdot \frac{\alpha - i}{\alpha - i} = \frac{(2 - i)(\alpha - i)}{(\alpha + i)(\alpha - i)} \] Simplifying the denominator: \[ (\alpha + i)(\alpha - i) = \alpha^2 + 1 \] Expanding the numerator: \[ (2 - i)(\alpha - i) = 2\alpha - 2i - i\alpha + i^2 = 2\alpha - i(\alpha + 2) - 1 \] Thus: \[ Z = \frac{2\alpha - 1 - i(\alpha + 2)}{\alpha^2 + 1} \] Now, the real and imaginary parts of \( Z \) are: \[ \text{Re}(Z) = \frac{2\alpha - 1}{\alpha^2 + 1}, \quad \text{Im}(Z) = \frac{-(\alpha + 2)}{\alpha^2 + 1} \]

2. Step 2: Use the given condition \( 4 \, \text{Re}(Z) = 3 \, \text{Im}( \overline{Z} ) \). The conjugate of \( Z \), \( \overline{Z} \), has the real part as \( \frac{2\alpha - 1}{\alpha^2 + 1} \) and the imaginary part as \( \frac{\alpha + 2}{\alpha^2 + 1} \). The condition \( 4 \, \text{Re}(Z) = 3 \, \text{Im}( \overline{Z} ) \) gives: \[ 4 \cdot \frac{2\alpha - 1}{\alpha^2 + 1} = 3 \cdot \frac{\alpha + 2}{\alpha^2 + 1} \] Simplifying: \[ 4(2\alpha - 1) = 3(\alpha + 2) \] Expanding: \[ 8\alpha - 4 = 3\alpha + 6 \] Solving for \( \alpha \): \[ 8\alpha - 3\alpha = 6 + 4 \quad \Rightarrow \quad 5\alpha = 10 \quad \Rightarrow \quad \alpha = 2 \] Thus, the value of \( \alpha \) is \( -2 \).