Question

The number $\frac {(1+i)^n} {(1-i)^{n-2}}$ is equal to

• A. 4i$^{n-2}$
• B. 2i$^{n-4}$
• C. 2i$^{n-1}$
• D. none of these

Answer 1

Answer: (C)

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

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

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

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