A) 1
B) - 1
C) \[i\]
D) \[-i\]
Correct Answer: C
Solution :
Since \[\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=i\] Therefore \[{{\left( \frac{1+i}{1-i} \right)}^{4n+1}}={{i}^{4n+1}}=i{{i}^{4n}}=i\,\,\,\,\,(\because {{i}^{4n}}=1)\]You need to login to perform this action.
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