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)

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