A) 2
B) 4
C) 8
D) None of these
Correct Answer: B
Solution :
\[\frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{{{(1+i)}^{2}}}{2}=\frac{2i}{2}=i\] \[\therefore \]\[{{\left( \frac{1+i}{1-i} \right)}^{m}}={{i}^{m}}=1\] (as given) So the least value of \[m=4\]\[\{\because {{i}^{4}}=1\}\]
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