Complex Numbers
Powers of Complex Numbers
Grade 11
Question:
<p>The least positive integer <span>\(n\)</span> for which <span>\(\left(\dfrac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)^n = 1\)</span>, is</p>
<p>2</p>
<p>3</p>
<p>5</p>
<p>6</p>
Step-by-Step Solution
Key Concept: Convert the complex number to polar form using De Moivre's theorem, then find the smallest positive n where the argument becomes a multiple of 2π.
<p><strong>Step 1:</strong> Simplify the complex fraction by rationalizing.</p><p>$$\frac{1+i\sqrt{3}}{1-i\sqrt{3}} = \frac{(1+i\sqrt{3})(1+i\sqrt{3})}{(1-i\sqrt{3})(1+i\sqrt{3})} = \frac{1+2i\sqrt{3}-3}{1+3} = \frac{-2+2i\sqrt{3}}{4} = \frac{-1+i\sqrt{3}}{2}$$</p><p><strong>Step 2:</strong> Convert to polar form. For $z = \frac{-1+i\sqrt{3}}{2}$:</p><p>$$|z| = \sqrt{\frac{1}{4}+\frac{3}{4}} = 1$$</p><p>$$\arg(z) = \pi - \arctan\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$</p><p>So $z = e^{i\frac{2\pi}{3}}$</p><p><strong>Step 3:</strong> Apply De Moivre's theorem:</p><p>$$z^n = e^{i\frac{2\pi n}{3}} = 1$$</p><p>This requires $\frac{2\pi n}{3} = 2\pi k$ for some integer $k$</p><p>$$n = 3k$$</p><p><strong>Step 4:</strong> The least positive integer is $n = 3$ (when $k=1$).</p><p>Verification: $z^3 = e^{i2\pi} = 1$ ✓</p><p>∴ Answer: B (n = 3)</p>
Correct Answer: B