The smallest positive integer $n$ for which $\left(\dfrac{1+i}{1-i}\right)^n = 1$ is
Step-by-Step Solution
Key Concept: Always simplify the base to a standard root of unity first; the multiplicative order of $i$ is 4, so $i^n = 1 \Leftrightarrow 4 \mid n$.
**Step 1: Simplify the base**
$\dfrac{1+i}{1-i} \cdot \dfrac{1+i}{1+i} = \dfrac{(1+i)^2}{2} = \dfrac{2i}{2} = i$.
**Step 2: Solve i^n = 1**
$i^n = 1$ iff $4 \mid n$. The smallest positive integer is $n = 4$.
**Step 3: Check options**
$n = 4$ does not appear among options (a) 8, (b) 16, (c) 12. So the answer is (d) None of these.
Correct Answer: 4