Question

The value of ${\left(\frac{1+i}{1-i}\right)}^{496}=$

Answer 1

$\left(\frac{1+i}{1-i}\right)=(\frac{1+i}{1-i}.\frac{1+i}{1+i})$
$=\frac{1+i^2+2i}{1-i^2}=\frac{1-1^2+2i}{2}=i$
Now
${\left(\frac{1+i}{1-i}\right)}^{496}=i^{496}=i^{4\times 124}$
We know that $i^4=1$
$\therefore i^{4\times 124}=1$