Question

The value of ${\left(\frac{1+i}{1-i}\right)}^{100}$ is equal to

• A. 1
• B. -1
• C. i
• D. -i

Answer 1

Answer: (A)

We have,
${\left(\frac{1+i}{1-i}\right)}^{100}={\left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]}^{100}$
$={\left[\frac{(1+i{)}^2}{1^2-i^2}\right]}^{100}$
$[\because (a-b)(a+b)=a^2-b^2]$
$={\left[\frac{(1+i{)}^2}{1+1}\right]}^{100}\left[\because i^2=-1\right]$
$={\left[\frac{1+i^2+2i}{2}\right]}^{100}$
$\left[\because (a+b{)}^2=a^2+b^2+2ab\right]$
$={\left[\frac{1-1+2i}{2}\right]}^{100}\left[\because 1^2=-1\right]$
$={\left[\frac{2i}{2}\right]}^{100}$
$=(i{)}^{100}={\left(i^2\right)}^{50}$
$=(-1{)}^{50}\left[\because 1^2=-1\right]$
$=1\left[\because (-1{)}^{\text{even }}=1\right]$