Question

The least positive integral value of $\text{m}$ , if ${\left(\frac{1+i}{1-i}\right)}^{\text{m}}=1$ is: $($ where $i=\sqrt{-1}$

Answer 1

Answer: 4

We have $\text{:}$ $\frac{1+\text{i}}{1-\text{i}}=\frac{1+\text{i}}{1-\text{i}}\times \frac{1+\text{i}}{1+\text{i}}=\frac{{\left(1+\text{i}\right)}^2}{1-{\left(\text{i}\right)}^2}$
$=\frac{1+2\text{i}+{\text{i}}^2}{1-{\text{i}}^2}$
$=\frac{1+2\text{i}-1}{1+1}=\frac{2\text{i}}{2}=\text{i}$
$\therefore {\left(\frac{1+\text{i}}{1-\text{i}}\right)}^{\text{m}}=1\Rightarrow {\left(\text{i}\right)}^{\text{m}}=1$
$\Rightarrow$ $\text{m}$ is a multiple of $4$ .
Hence, the smallest positive value of $\text{m}$ is $\text{4}$ .