Question

The smallest positive integer $n$, for which $\frac{(1+i{)}^n}{(1-i{)}^{n-2}}$ is a real number is

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

Answer: (C)

$\frac{{\left(1+i\right)}^n}{{\left(1-i\right)}^{n-2}}={\left(\frac{1+i}{1-i}\right)}^n{\left(1-i\right)}^2$
$={\left(\frac{1+i}{1-i}\cdot \frac{1+i}{1+i}\right)}^n\left[1+i^2-2i\right]$
$={\left(\frac{1+2i+i^2}{1-i^2}\right)}^n\left(-2i\right)={\left(i\right)}^n\left(-2i\right)$
$=-2{\left(i\right)}^{n+1}$ which is real if
$n+1=2$ i.e., $n=1$ (Here $n$ is smallest $+ve$ integer)