Question

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

Answer 1

${\left(\frac{1+i}{1-i}\right)}^{496}={\left[{\left(\frac{1+i}{1-i}\right)}^2\right]}^{248}\Rightarrow {\left(\frac{1+i^2+2i}{1+i^2-2i}\right)}^{248}\Rightarrow {\left(\frac{1-1+2i}{1-1-2i}\right)}^{248}\Rightarrow {\left(\frac{2i}{-2i}\right)}^{248}\Rightarrow (-1{)}^{248}=1$