Question

Let \( z = \frac{2 - i}{\alpha + i} \), where \( \alpha \) is a real number. If \( 4\text{Re}(z) = 3\text{Im}(\bar{z}) \), then the value of \( \alpha \) is

• A. \(5 \)
• B. \(-5 \)
• C. \(3 \)
• D. \(2 \)
• E. \(-2 \)

Hint:

Always rationalize complex fractions first, then separate real and imaginary parts.

Answer 1

The Correct Option is D

Concept: \[ \text{Im}(\bar{z}) = -\text{Im}(z) \]

Step 1: Rationalize denominator.
\[ z = \frac{2 - i}{\alpha + i} \cdot \frac{\alpha - i}{\alpha - i} \] \[ = \frac{(2 - i)(\alpha - i)}{\alpha^2 + 1} \]

Step 2: Expand numerator.
\[ (2 - i)(\alpha - i) = 2\alpha - 2i - \alpha i + i^2 \] \[ = 2\alpha - (2 + \alpha)i -1 \] \[ = (2\alpha - 1) - (2 + \alpha)i \]

Step 3: Extract real and imaginary parts.
\[ \text{Re}(z) = \frac{2\alpha - 1}{\alpha^2 + 1}, \quad \text{Im}(z) = \frac{-(2 + \alpha)}{\alpha^2 + 1} \] \[ \text{Im}(\bar{z}) = \frac{2 + \alpha}{\alpha^2 + 1} \]

Step 4: Apply given condition.
\[ 4 \cdot \frac{2\alpha - 1}{\alpha^2 + 1} = 3 \cdot \frac{2 + \alpha}{\alpha^2 + 1} \]

Step 5: Solve equation.
\[ 4(2\alpha - 1) = 3(2 + \alpha) \] \[ 8\alpha - 4 = 6 + 3\alpha \] \[ 5\alpha = 10 \Rightarrow \alpha = 2 \]