Question
If $${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1\,\,{\text{then}}$$
A.
$$x = 2n + 1 ,$$ 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
Answer :
$$x = 4n ,$$ where $$n$$ is any positive integer
Solution :
$$\eqalign{
& {\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1 \cr
& \Rightarrow \,\,{\left[ {\frac{{{{\left( {1 + i} \right)}^2}}}{{1 - {i^2}}}} \right]^x} = 1 \cr
& \left( {\frac{{1 + {i^2} + 2i}}{{1 + 1}}} \right) = 1 \cr
& \Rightarrow \,\,{\left( i \right)^x} = 1; \cr
& \therefore \,\,x = 4n;\,\,\,\,n \in {I^ + } \cr} $$