Question

If $\left( \frac{1 - i}{1 + i} \right)^{100} = a + ib$, then

• A. a = 2, b = –1
• B. a = 1, b = 0
• C. a = 0, b = 1
• D. a = –1, b = 2

Answer 1

Answer: (B)

a = 1, b = 0

Answer 2

Sol.$\frac{\mathbf{1 - i}}{\mathbf{1 + i}}\mathbf{\times}\frac{\mathbf{1 - i}}{\mathbf{1 - i}}\mathbf{= - i =}\mathbf{\cos}\left( \mathbf{-}\frac{\mathbf{\pi}}{\mathbf{2}} \right)\mathbf{+ i}\mathbf{\sin}\left( \mathbf{-}\frac{\mathbf{\pi}}{\mathbf{2}} \right)\mathbf{\Rightarrow ( - i}\mathbf{)}^{\mathbf{100}}\mathbf{=}\mathbf{\cos}\mathbf{(}\mathbf{- 50\pi) + i}\mathbf{\sin}\mathbf{(}\mathbf{- 50\pi)}$ = $\mathbf{1 + i(0)}$ $\mathbf{\Rightarrow}\mathbf{a = 1,b = 0}$