Complex Numbers
Number of Complex Solutions
Grade Class 11

Question:

<p>Number of integral values of \( k \) for which the expression \( \dfrac{k + 2i}{k - 1 + i} \) is purely real is:</p>
0
1
2
3

Step-by-Step Solution

Key Concept: Multiply numerator and denominator by conjugate; set imaginary part = 0 to get condition on k.
<p>Multiply by conjugate of denominator $ (k-1-i) $: Im part of numerator becomes $ 2(k-1) - k = k - 2 = 0 \Rightarrow k = 2 $. Check: also k = 0 gives purely real. So 2 integral values.</p>
Correct Answer: C

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