Question

The value of \( \alpha \) for which the complex number \( \frac{2 - \alpha i}{\alpha - i} \) is purely imaginary, is:

• A. 2
• B. -2
• C. 1
• D. -1
• E. 0

Hint:

To make a complex number purely imaginary, set the real part of the expression equal to zero and solve for the unknown variable.

Answer 1

Answer: (E)

Let the complex number be \( z = \frac{2 - \alpha i}{\alpha - i} \).
To determine the value of \( \alpha \) such that \( z \) is purely imaginary, we must eliminate the real part of the complex number.
First, multiply both the numerator and denominator by the complex conjugate of the denominator \( \alpha + i \):
\[ z = \frac{(2 - \alpha i)(\alpha + i)}{(\alpha - i)(\alpha + i)}. \] The denominator simplifies using the difference of squares: \[ (\alpha - i)(\alpha + i) = \alpha^2 + 1. \] Now, expand the numerator: \[ (2 - \alpha i)(\alpha + i) = 2\alpha + 2i - \alpha^2 i - \alpha i^2. \] Since \( i^2 = -1 \), this becomes: \[ 2\alpha + 2i - \alpha^2 i + \alpha. \] Now group the real and imaginary parts: \[ {Real part: } 2\alpha + \alpha = 3\alpha, \quad {Imaginary part: } 2 - \alpha^2. \] Thus, we have: \[ z = \frac{3\alpha + (2 - \alpha^2)i}{\alpha^2 + 1}. \] For \( z \) to be purely imaginary, the real part must be 0: \[ 3\alpha = 0 \quad \Rightarrow \quad \alpha = 0. \] Thus, the value of \( \alpha \) for which \( z \) is purely imaginary is \( \alpha = 0 \).