<p>Find the least positive integer \(n\) such that \(\left(\dfrac{2i}{1+i}\right)^n\) is a positive integer.</p>
Step-by-Step Solution
Key Concept: Simplify the complex fraction by rationalizing, then express in polar form to find when the result is a positive real integer by analyzing the argument and modulus.
<p><strong>Step 1: Simplify the complex fraction</strong></p><p>$$\frac{2i}{1+i} = \frac{2i}{1+i} \cdot \frac{1-i}{1-i} = \frac{2i(1-i)}{(1+i)(1-i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i + 2}{1 + 1} = \frac{2 + 2i}{2} = 1 + i$$</p><p><strong>Step 2: Convert to polar form</strong></p><p>$$1 + i = \sqrt{2}e^{i\pi/4} = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$$</p><p><strong>Step 3: Apply De Moivre's theorem</strong></p><p>$$(1+i)^n = (\sqrt{2})^n e^{in\pi/4} = 2^{n/2}\left(\cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4}\right)$$</p><p><strong>Step 4: Find conditions for positive integer</strong></p><p>For this to be a positive integer:</p><p>• Imaginary part = 0: $\sin\frac{n\pi}{4} = 0 \Rightarrow n\pi/4 = k\pi \Rightarrow n = 4k$</p><p>• Real part positive: $\cos\frac{n\pi}{4} > 0 \Rightarrow n = 4k$ where $k$ is odd (so $n = 4, 12, 20, ...$) gives $\cos 0 = 1$ when $n \equiv 0 \pmod{8}$</p><p>• Modulus is integer: $2^{n/2} \in \mathbb{Z} \Rightarrow n/2 \in \mathbb{Z} \Rightarrow n$ is even</p><p><strong>Step 5: Check smallest values</strong></p><p>For $n = 4$: $(1+i)^4 = 2^2(\cos\pi + i\sin\pi) = 4(-1) = -4$ (negative)</p><p>For $n = 8$: $(1+i)^8 = 2^4(\cos 2\pi + i\sin 2\pi) = 16(1) = 16$ (positive integer ✓)</p><p>∴ Answer: <strong>8</strong></p>
Correct Answer: 8