<p>If \ \(\left(\dfrac{1+i}{1-i}\right)^m = 1\), then find the least positive integral value of \ \(m\).</p>
Step-by-Step Solution
Key Concept: Simplify the complex fraction first by multiplying by the conjugate, then use De Moivre's theorem: if z^m = 1, find the smallest positive integer m where the argument cycles back to 0 (or 2πk).
<p><strong>Step 1:</strong> Simplify $\frac{1+i}{1-i}$ by multiplying numerator and denominator by the conjugate $(1+i)$:</p><p>$$\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2}$$</p><p><strong>Step 2:</strong> Evaluate using $i^2 = -1$:</p><p>$$= \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$$</p><p><strong>Step 3:</strong> Now solve $i^m = 1$. Since $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$, we find that $i^m = 1$ when $m = 4k$ for positive integers $k$.</p><p><strong>Step 4:</strong> The least positive integral value is when $k = 1$.</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4