<p>If the six solutions of <span>\(x^6 = -64\)</span> are written in the form <span>\(a + bi\)</span>, where <span>\(a\)</span> and <span>\(b\)</span> are real, then find the product of those solutions with <span>\(a > 0\)</span>.</p>
Step-by-Step Solution
Key Concept: Express -64 in polar form as 64e^(iπ), then use De Moivre's theorem to find all six sixth roots. Identify which roots have positive real parts and multiply them.
<p><strong>Step 1:</strong> Express -64 in polar form: -64 = 64e^(i(π+2πk)) where k = 0,1,2,3,4,5</p><p><strong>Step 2:</strong> Apply De Moivre's theorem. The six sixth roots are: x = 2e^(i(π+2πk)/6) for k = 0,1,2,3,4,5</p><p>These give angles: π/6, π/2, 5π/6, 7π/6, 3π/2, 11π/6</p><p><strong>Step 3:</strong> Convert to rectangular form:</p><p>• k=0: 2e^(iπ/6) = 2(cos π/6 + i sin π/6) = √3 + i (a > 0) ✓</p><p>• k=1: 2e^(iπ/2) = 2i (a = 0) ✗</p><p>• k=2: 2e^(i5π/6) = 2(cos 5π/6 + i sin 5π/6) = -√3 + i (a < 0) ✗</p><p>• k=3: 2e^(i7π/6) = 2(cos 7π/6 + i sin 7π/6) = -√3 - i (a < 0) ✗</p><p>• k=4: 2e^(i3π/2) = -2i (a = 0) ✗</p><p>• k=5: 2e^(i11π/6) = 2(cos 11π/6 + i sin 11π/6) = √3 - i (a > 0) ✓</p><p><strong>Step 4:</strong> Multiply the roots with a > 0:</p><p>(√3 + i)(√3 - i) = (√3)² - (i)² = 3 - (-1) = 3 + 1 = 4</p><p>∴ Answer: 4</p>
Correct Answer: 4