• # question_answer If  $n$ is a positive integer,  then ${{\left( \frac{1+i}{1-i} \right)}^{4n+1}}$= A) 1 B) - 1 C) $i$ D) $-i$

Since $\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=i$ Therefore ${{\left( \frac{1+i}{1-i} \right)}^{4n+1}}={{i}^{4n+1}}=i{{i}^{4n}}=i\,\,\,\,\,(\because {{i}^{4n}}=1)$