<p>The values of <span class="math">(16)^{1/4}</span> are</p>
<p>(a) <span class="math">\pm 2, \pm 2i</span></p>
<p>(b) <span class="math">\pm 4, \pm 4i</span></p>
<p>(c) <span class="math">\pm 1, \pm i</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: To find all fourth roots of 16, we convert to polar form and apply De Moivre's theorem: if z = r(cos θ + i sin θ), then z^(1/n) has n distinct values given by r^(1/n)[cos((θ + 2πk)/n) + i sin((θ + 2πk)/n)] for k = 0, 1, ..., n-1.
<p><strong>Step 1:</strong> Express 16 in polar form.</p><p>16 = 16(cos 0° + i sin 0°) = 16e^(i·0)</p><p>More generally: 16 = 16(cos 2πk + i sin 2πk) for k = 0, 1, 2, 3, ...</p><p><strong>Step 2:</strong> Apply De Moivre's theorem to find the fourth roots.</p><p>If w = (16)^(1/4), then w = 16^(1/4)[cos((2πk)/4) + i sin((2πk)/4)]</p><p>where k = 0, 1, 2, 3 (giving 4 distinct roots)</p><p><strong>Step 3:</strong> Calculate 16^(1/4) = (2^4)^(1/4) = 2</p><p><strong>Step 4:</strong> Find all four roots by substituting k = 0, 1, 2, 3:</p><p>• k = 0: w = 2[cos(0°) + i sin(0°)] = 2(1 + 0i) = <strong>2</strong></p><p>• k = 1: w = 2[cos(π/2) + i sin(π/2)] = 2(0 + i) = <strong>2i</strong></p><p>• k = 2: w = 2[cos(π) + i sin(π)] = 2(-1 + 0i) = <strong>-2</strong></p><p>• k = 3: w = 2[cos(3π/2) + i sin(3π/2)] = 2(0 - i) = <strong>-2i</strong></p><p><strong>Step 5:</strong> Verify each root by raising to the 4th power:</p><p>(±2)^4 = 16 ✓</p><p>(±2i)^4 = (2i)^4 = 2^4·i^4 = 16·1 = 16 ✓</p><p><strong>∴ Answer:</strong> A</p>
Correct Answer: A