Question

Find the smallest positive integer value of $n$ for which $\frac{(1+i{)}^n}{(1-i{)}^{n-2}}$ is a real number.

Answer 1

Now,

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

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

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

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

$=i^n.(-2i)$

For the given expression to be real for smallest integral value of $n$ we must have $n=1$.

Then the value of the expression will be $2$.