Complex Numbers
Roots of complex equations
Grade None
Question:
<p>Consider the equation \(10z^2 - 3iz - k = 0\), where \(z\) is a complex variable and \(i^2 = -1\). Which of the following statements is true?</p>
<p>For real positive numbers \(k\), both roots are purely imaginary.</p>
<p>For all complex numbers \(k\), neither root is real.</p>
<p>For all purely imaginary numbers \(k\), both roots are real and irrational.</p>
<p>For real negative numbers \(k\), both roots are purely imaginary.</p>
Step-by-Step Solution
Key Concept: For a quadratic with real coefficient k and complex coefficients, if one root is purely imaginary (z = bi), substitute directly to find k, then verify the discriminant and root properties.
<p><strong>Step 1:</strong> Assume one root is purely imaginary, say z = bi where b ∈ ℝ.</p><p><strong>Step 2:</strong> Substitute z = bi into 10z² - 3iz - k = 0:</p><p>10(bi)² - 3i(bi) - k = 0</p><p>10(-b²) - 3b·i² - k = 0</p><p>-10b² - 3b(-1) - k = 0</p><p>-10b² + 3b - k = 0</p><p><strong>Step 3:</strong> For this to hold for real b and k, separate into real and imaginary parts. Since all terms are real, we need: k = -10b² + 3b</p><p><strong>Step 4:</strong> Use the quadratic formula: z = [3i ± √(9i² + 40k)]/20 = [3i ± √(-9 + 40k)]/20</p><p>For purely imaginary roots, the discriminant -9 + 40k must yield a real value. If k = 0: z = [3i ± 3i]/20, giving z = 3i/10 or z = 0. Testing z = 3i/10: 10(9i²/100) - 3i(3i/10) = -9/10 + 9/10 = 0 ✓</p><p>∴ Answer: A</p>
Correct Answer: A