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