Complex Numbers
Powers of Complex Numbers
Grade 11
Question:
<p>If \(\left(\dfrac{1+i}{1-i}\right)^x = 1\), then</p>
<p>\(x = 4n\), where \(n\) is any positive integer.</p>
<p>\(x = 2n\), where \(n\) is any positive integer.</p>
<p>\(x = 4n + 1\), where \(n\) is any positive integer.</p>
<p>\(x = 2n + 1\), where \(n\) is any positive integer.</p>
Step-by-Step Solution
Key Concept: Convert the complex fraction to standard form by rationalizing (multiply by conjugate), then use the fact that a complex number equals 1 only when it's a real number equal to 1, or recognize the pattern of powers of i.
<p><strong>Step 1:</strong> Simplify the complex fraction by multiplying numerator and denominator by the conjugate of the denominator:</p><p>$$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$$</p><p><strong>Step 2:</strong> The equation becomes $i^x = 1$</p><p><strong>Step 3:</strong> Recall the pattern of powers of i: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, $i^5=i$, ...</p><p><strong>Step 4:</strong> Since $i^x = 1$, we need $x = 4n$ where $n$ is any non-negative integer.</p><p>∴ Answer: x must be a multiple of 4 (or $x = 4n, n \in \mathbb{Z}$)</p>
Correct Answer: A