If $(\left(\frac{1+i}{1-i}\right)^x = 1)$, then
$(x = 4n + 1; \, n \in \mathbb{N})$
$(x = 2n + 1; \, n \in \mathbb{N})$
$(x = 2n; \, n \in \mathbb{N})$
$(x = 4n; \, n \in \mathbb{N})$
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