If $\left(\dfrac{1+i}{1-i}\right)^x = 1$, then A). $x = 4n+1$, where $n$ is any positive integer B). $x = 4n$, where $n$ is any positive integer C). $x = 2n$, where $n$ is any positive integer D). $x = 4n+1$, where $n$ is any positive integer

The correct answer is $x = 4n$, where $n$ is any positive integer. **Solution:** $$\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1+2i-1}{2} = i$$ So $(i)^x = 1$. Since $i^4 = 1$, we need $x = 4n$ where $n \in \mathbb{Z}^+$.

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