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)

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

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

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

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