Complex Numbers
Powers of Complex Numbers
Grade 11
Question:
<p>If \(\left(\dfrac{1+i}{1-i}\right)^x = 1\), then \(x\) is:</p>
<p>\(x = 2n+1, n \in \mathbb{Z}\)</p>
<p>\(x = 4n, n \in \mathbb{Z}\)</p>
<p>\(x = 2n, n \in \mathbb{Z}\)</p>
<p>\(x = n, n \in \mathbb{Z}\)</p>
Step-by-Step Solution
Key Concept: Simplify the complex fraction by multiplying by the conjugate, then recognize that i² = -1 gives i, and find the smallest positive integer where i^x = 1.
<p><strong>Step 1:</strong> Simplify the complex fraction by multiplying by conjugate of denominator.</p><p>$$\frac{1+i}{1-i} = \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$$</p><p><strong>Step 2:</strong> Expand numerator and denominator.</p><p>Numerator: $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$</p><p>Denominator: $(1-i)(1+i) = 1 - i^2 = 1 - (-1) = 2$</p><p><strong>Step 3:</strong> Simplify the fraction.</p><p>$$\frac{1+i}{1-i} = \frac{2i}{2} = i$$</p><p><strong>Step 4:</strong> Solve $i^x = 1$ by using powers of $i$.</p><p>$i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ ✓</p><p>The pattern repeats with period 4.</p><p><strong>Step 5:</strong> Determine possible values.</p><p>$i^x = 1$ when $x = 4, 8, 12, ...$ or $x = 4n$ where $n \in \mathbb{Z}^+$</p><p>∴ Answer: B (assuming B represents $x = 4n$ or the smallest value $x = 4$)</p>
Correct Answer: B