Question

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

Answer 1

Consider $\left(\frac{1+i}{1-i}\right)$

Rationalize the denominator we get,

$\Rightarrow \left(\frac{1+i}{1-i}\right)=\frac{(1+i)(1+i)}{(1-i)(1+1)}=\frac{(1+i{)}^2}{(1-i)(1+i)}$

Here we see that denominator is in the form of $(a+b)(a-b)=a^2-b^2$

$\Rightarrow \frac{1+i}{1-i}=\frac{(1+i{)}^2}{1^2-i^2}$

$\Rightarrow \frac{1+i}{1-i}=\frac{(1+i{)}^2}{1-i^2}$

we know that $(a+b{)}^2=a^2+2ab+b^2$

$\Rightarrow \frac{1+i}{1-i}=\frac{1^2+(2\times 1\times i)+i^2}{1-i^2}$

$\Rightarrow \frac{1+i}{1-i}=\frac{1^2+2i+i^2}{1-i^2}$

We know that $i^2=-1$ , substituting this we get,

$\Rightarrow \frac{1+i}{1-i}=\frac{1+2i+(-1)}{1-(-1)}$

$\Rightarrow \frac{1+i}{1-i}=\frac{1+2i-1}{1+1}$

$\Rightarrow \frac{1+i}{1-i}=\frac{2i}{2}$

$\Rightarrow \frac{1+i}{1-i}=i$

Given that $\Rightarrow {\left(\frac{1+i}{1-i}\right)}^m=1$

$\Rightarrow {\left(\frac{1+i}{1-i}\right)}^m=i^m$

So, $i^m=1$

We know that $i^2=-1$

Squaring on both sides we get,

$i^4=(-1{)}^2$

$i^4=1$

Therefore the smallest value of $m$ for which ${\left(\frac{1+i}{1-i}\right)}^m=1$ is $m=4$