Question

If ${\left(\frac{1+i}{1-i}\right)}^n$ is 1 find the least value of n where $n\in N$.

Answer 1

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

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

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

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

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

$={\left(\frac{2i}{1+1}\right)}^n=i^n=$Imaginary

$\Rightarrow n=1,3,5,...$ for positive imaginary part $n=3$