# Solve the following

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.