Complex Numbers
Roots of Complex Numbers
Grade 11
Question:
<p>Least positive argument of the 4th root of the complex number <math>2 - i\sqrt{12}</math> is:</p>
<p>(a) <math>\frac{\pi}{6}</math></p>
<p>(b) <math>\frac{\pi}{12}</math></p>
<p>(c) <math>\frac{5\pi}{12}</math></p>
<p>(d) <math>\frac{7\pi}{12}</math></p>
Step-by-Step Solution
Key Concept: Convert to polar form, then use the formula for nth roots which give arguments differing by <math>\frac{2\pi}{n}</math>.
<p>Express <math>2 - i\sqrt{12}</math> in polar form.</p><p><math>2 - i\sqrt{12} = 2 - 2i\sqrt{3}</math></p><p><math>|2 - 2i\sqrt{3}| = \sqrt{4 + 12} = 4</math></p><p><math>\arg(2 - 2i\sqrt{3}) = -\frac{\pi}{3}</math></p><p>The 4th roots have arguments: <math>-\frac{\pi}{12}, \frac{5\pi}{12}, \frac{11\pi}{12}, \frac{17\pi}{12}</math></p><p>The least positive argument is <math>\frac{5\pi}{12}</math>.</p>
Correct Answer: C