Question

If $\left( \frac{1 + i}{1 - i} \right)^m =1$, then the least positive integral value of $m$ is

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

Answer: (C)

4

Answer 2

We have,
$\left(\frac{1+i}{1-i}\right)^{m}=1$
$\Rightarrow \left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^{m}=1$
$\Rightarrow \left[\frac{1+2 i+i^{2}}{1-i^{2}}\right]^{m}=1$
$\Rightarrow \left[\frac{1+2 i-1}{1-(-1)}\right]^{m}=1 \left[\because i^{2}=-1\right]$
$\Rightarrow \left[\frac{2 i}{2}\right]^{m}=1$
$\Rightarrow i^{m}=1$
$\Rightarrow i^{m}=1^{4} [\because i^4 = 1]$
$\Rightarrow m=4$