<p>Given \(z\) is a complex number with modulus 1. Then the equation \([(1 + ia)/(1 - ia)]^4 = z\) has</p>
<p>all roots real and distinct</p>
<p>two real and two imaginary</p>
<p>three roots real and one imaginary</p>
<p>one root real and three imaginary</p>
Step-by-Step Solution
Key Concept: Since |z| = 1 and z = [(1+ia)/(1-ia)]^4, we must have |[(1+ia)/(1-ia)]^4| = 1. This means |(1+ia)/(1-ia)| = 1, which implies |1+ia| = |1-ia|, constraining the parameter a. The number of solutions depends on how many values of a satisfy this geometric constraint.
<p><strong>Step 1:</strong> Since z has modulus 1, we have |z| = 1.</p><p><strong>Step 2:</strong> Taking modulus of both sides: |[(1+ia)/(1-ia)]^4| = |z| = 1</p><p>This gives: |(1+ia)/(1-ia)|^4 = 1, so |(1+ia)/(1-ia)| = 1</p><p><strong>Step 3:</strong> Computing the modulus of the fraction:</p><p>|(1+ia)/(1-ia)| = |1+ia|/|1-ia| = √(1+a²)/√(1+a²) = 1 ✓</p><p>This is satisfied for ALL real values of a.</p><p><strong>Step 4:</strong> For any real a, [(1+ia)/(1-ia)]^4 lies on the unit circle. Writing (1+ia)/(1-ia) in exponential form: it equals e^(i·2arctan(a))</p><p>Therefore: [(1+ia)/(1-ia)]^4 = e^(i·8arctan(a))</p><p><strong>Step 5:</strong> As a ranges over all real numbers, 8arctan(a) ranges over (-4π, 4π), allowing the fourth power to represent any point on the unit circle exactly once in the principal domain.</p><p>∴ Answer: A (The equation has a unique solution, or infinitely many solutions depending on whether a is fixed or variable)</p>
Correct Answer: A