Question:
If $\left(\frac{1+i}{1-i}\right)^{n}=1$, then $n=$
(a) 2m + 1
(b) 4m
(c) 2m
(d) 4m + 1 where m ∈ N
Solution:
Given :- $\left(\frac{1+i}{1-i}\right)^{n}=1$
Since, $\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$
$=\frac{(1+i)^{2}}{1-i^{2}}$
$=\frac{1+i^{2}+2 i}{1+1}$
i. e $\frac{1+i}{1-i}=\frac{1-1+2 i}{2}=\frac{2 i}{2}=2 i$
$\Rightarrow\left(\frac{1+i}{1-i}\right)^{n}=1$
$\Rightarrow(i)^{n}=1$
$\Rightarrow n=4 \mathrm{~m} \quad$ where $m \in N$
Hence, the correct answer is option B.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.