Question

The expression $\frac{\left(1+i\right)^{n}}{\left(1-i\right)^{n-2}}$ equals

• A. $-i^{n+1}$
• B. $i^{n+1}$
• C. $-2i^{n+1}$
• D. $1$

Answer 1

Answer: (C)

$\frac{(1+i)^{n}}{(1-i)^{n-2}}=\frac{(1+i)^{n}}{(1-i)^{n}(1-i)^{-2}}$
$=\left(\frac{1+i}{1-i}\right)^{n}(1-i)^{2}$
$=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{n}\left(1+i^{2}-2 i\right)$
$=\left(\frac{1+2 i+i^{2}}{1-i^{2}}\right)^{n}\left(1+i^{2}-2 i\right)$
$=\left(\frac{1+2 i-1}{1-(-1)}\right)^{n}(1-1-2 i)$
$\left[\because i^{2}=-1\right]$
$=\left(\frac{2 i}{2}\right)^{n}(-2 i)$
$=i^{n}(-2 i)=-2 i^{n+1}$