<p>If \(a^2 + b^2 = 1\), then \(\dfrac{1 + b + ia}{1 + b - ia} =\)</p>
Step-by-Step Solution
Key Concept: Multiply numerator and denominator by the conjugate of the denominator to simplify the complex fraction, then use the constraint a² + b² = 1 to reduce the result to a clean form.
<p><strong>Step 1:</strong> Multiply numerator and denominator by the conjugate of denominator (1 + b + ia):</p><p>$$\frac{1+b+ia}{1+b-ia} \cdot \frac{1+b+ia}{1+b+ia} = \frac{(1+b+ia)^2}{(1+b)^2 + a^2}$$</p><p><strong>Step 2:</strong> Expand numerator: $(1+b+ia)^2 = (1+b)^2 + 2ia(1+b) + (ia)^2 = (1+b)^2 - a^2 + 2ia(1+b)$</p><p><strong>Step 3:</strong> Expand denominator: $(1+b)^2 + a^2 = 1 + 2b + b^2 + a^2 = 1 + 2b + 1 = 2(1+b)$ (using $a^2 + b^2 = 1$)</p><p><strong>Step 4:</strong> Numerator becomes: $(1+b)^2 - a^2 + 2ia(1+b) = (1+b)^2 - (1-b^2) + 2ia(1+b) = 2b(1+b) + 2ia(1+b) = 2(1+b)(b+ia)$</p><p><strong>Step 5:</strong> Simplify: $$\frac{2(1+b)(b+ia)}{2(1+b)} = b + ia$$</p><p>∴ Answer: <strong>b + ia</strong></p>
Correct Answer: C