<p>Find the principal argument of <strong>(c)</strong> \(\sin\alpha + i(1-\cos\alpha)\), \(0 < \alpha < \pi\)</p>
Step-by-Step Solution
Key Concept: Express the complex number in terms of half-angle formulas: sin α = 2sin(α/2)cos(α/2) and 1 - cos α = 2sin²(α/2), then factor out common terms to identify the argument directly.
<p><strong>Step 1:</strong> Use half-angle identities to rewrite the complex number.</p><p>sin α = 2sin(α/2)cos(α/2)</p><p>1 - cos α = 2sin²(α/2)</p><p><strong>Step 2:</strong> Substitute these into the complex number:</p><p>z = 2sin(α/2)cos(α/2) + i·2sin²(α/2)</p><p><strong>Step 3:</strong> Factor out 2sin(α/2):</p><p>z = 2sin(α/2)[cos(α/2) + i·sin(α/2)]</p><p><strong>Step 4:</strong> Recognize that cos(α/2) + i·sin(α/2) = e^(i·α/2)</p><p>Since 0 < α < π, we have 0 < α/2 < π/2, so sin(α/2) > 0.</p><p>The factor 2sin(α/2) is a positive real number (doesn't affect argument).</p><p><strong>Step 5:</strong> The principal argument of e^(i·α/2) is α/2.</p><p>∴ Answer: <strong>α/2</strong></p>
Correct Answer: α/2