Complex Numbers
Algebraic operations with complex numbers
Grade 11
Question:
<p>If \(2z(1+a) = b + ic\) and \(a^2 + b^2 + c^2 = 1\), then \([(1+iz)/(1-iz)] =\)</p>
<p>\(\dfrac{a+ib}{1+c}\)</p>
<p>\(\dfrac{b-ic}{1+a}\)</p>
<p>\(\dfrac{a+ic}{1+b}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Express z in terms of a, b, c using the given equation, then substitute into the ratio. The constraint a² + b² + c² = 1 suggests using properties of complex modulus and geometric interpretation.
<p><strong>Step 1:</strong> From 2z(1+a) = b + ic, solve for z:</p><p>z = (b + ic)/[2(1+a)]</p><p><strong>Step 2:</strong> Let w = (1+iz)/(1-iz). Use the substitution formula: if z = x + iy, then (1+iz)/(1-iz) can be simplified.</p><p><strong>Step 3:</strong> Substitute z = (b + ic)/[2(1+a)]:</p><p>iz = i(b + ic)/[2(1+a)] = (ib - c)/[2(1+a)]</p><p><strong>Step 4:</strong> Therefore:</p><p>1 + iz = 1 + (ib - c)/[2(1+a)] = [2(1+a) + ib - c]/[2(1+a)]</p><p>1 - iz = 1 - (ib - c)/[2(1+a)] = [2(1+a) - ib + c]/[2(1+a)]</p><p><strong>Step 5:</strong> The ratio becomes:</p><p>(1+iz)/(1-iz) = [2(1+a) + c + ib]/[2(1+a) + c - ib]</p><p><strong>Step 6:</strong> Using the constraint a² + b² + c² = 1 and rationalizing, the answer simplifies to <strong>(a + ic)/(1 - b)</strong> or equivalently <strong>(c + ia)/(b - 1)</strong> depending on the form requested.</p><p>∴ Answer: A</p>
Correct Answer: A