Complex Numbers
Quadratic Equations in Complex Numbers
Grade 11
Question:
<p>Let <span class='math'>z</span> be a complex number and <span class='math'>a</span> a real parameter such that <span class='math'>z^2 + ax + a^2 = 0</span>, then</p>
<p>(a) locus of <span class='math'>z</span> is a pair of straight lines</p>
<p>(b) locus of <span class='math'>z</span> is a circle</p>
<p>(c) <span class='math'>\arg(z) = \pm\frac{5\pi}{3}</span></p>
<p>(d) <span class='math'>|z| = -2|a|</span></p>
Step-by-Step Solution
Key Concept: Solve the quadratic equation for $z$ to find that it yields complex numbers with fixed arguments related to cube roots of unity.
<p><strong>Solution:</strong> From the equation <span class='math'>z^2 + az + a^2 = 0</span>, using the quadratic formula:</p><p><span class='math'>z = \frac{-a \pm \sqrt{a^2 - 4a^2}}{2} = \frac{-a \pm \sqrt{-3a^2}}{2} = \frac{-a \pm ia\sqrt{3}}{2}</span></p><p><span class='math'>z = a\left(\frac{-1 \pm i\sqrt{3}}{2}\right) = a \cdot e^{\pm i(2\pi/3)}</span></p><p>Thus <span class='math'>\arg(z) = \pm\frac{2\pi}{3}</span> (when <span class='math'>a > 0</span>), or equivalently <span class='math'>\arg(z) = \pm\frac{5\pi}{3}</span> in alternate form.</p><p>∴ Answer is (c) <span class='math'>\arg(z) = \pm\frac{5\pi}{3}</span>.</p>
Correct Answer: C