Question

Let \(z = \frac{a - \frac{i}{2}}{i - 2}\), where \(a\) is a real number and \(i = \sqrt{-1}\). If \(\text{Im}(z) = 0\), then the value of \(a\) is equal to

• A. 1
• B. 2
• C. 3
• D. 4
• E. 0

Hint:

If \(\frac{z_1}{z_2}\) is purely real, then \(z_1 \overline{z_2}\) is also purely real. You only need to set the imaginary part of the product of the numerator and the conjugate of the denominator to zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the imaginary part of a complex number in fractional form, we must express it in the standard form \(x + iy\).
This involves multiplying the numerator and denominator by the complex conjugate of the denominator.

Step 2: Key Formula or Approach:
For a complex number \(z = \frac{z_1}{z_2}\), multiply by \(\frac{\overline{z_2}}{\overline{z_2}}\).
Here, \(z = \frac{a - i/2}{-2 + i}\). The conjugate of the denominator \(-2 + i\) is \(-2 - i\).

Step 3: Detailed Explanation:
Rewrite \(z\):
\[ z = \frac{2a - i}{2(-2 + i)} \]
Rationalize the denominator:
\[ z = \frac{(2a - i)(-2 - i)}{2(-2 + i)(-2 - i)} \]
Expand the numerator:
\[ (2a - i)(-2 - i) = -4a - 2ai + 2i + i^2 \]
Since \(i^2 = -1\):
\[ = (-4a - 1) + i(2 - 2a) \]
Expand the denominator:
\[ 2[(-2)^2 + 1^2] = 2[4 + 1] = 10 \]
So, \(z = \frac{-4a - 1}{10} + i\frac{2 - 2a}{10}\).
Given \(\text{Im}(z) = 0\):
\[ \frac{2 - 2a}{10} = 0 \]
\[ 2 - 2a = 0 \implies a = 1 \]

Step 4: Final Answer:
The value of \(a\) is 1.