Question

If $[\frac{(1-i)}{(1+i)}{]}^{100}=a+ib,$ then values of $a,b$

Answer 1

Finding the values of $\mathit{a}\mathbf{,}\mathit{b}\mathbf{}$:

Given that ${\left[\frac{(1-i)}{(1+i)}\right]}^{100}=a+ib,$

$\frac{(1-i)}{(1+i)}=\frac{(1-i)(1-i)}{(1+i)(1-i)}$

$$\begin{gathered} =\frac{(1-2i-1)}{(1+1)} \\ =-\frac{2i}{2} \\ =-i \end{gathered}$$

${\left[\frac{(1-i)}{(1+i)}\right]}^{100}={\left(-i\right)}^{100}$

$$\begin{gathered} ={\left(i^4\right)}^{25} \\ =1 \end{gathered}$$

Comparing with $a+ib,a=1,b=0$

Hence, the values are $\mathit{a}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{b}\mathbf{=}\mathbf{0}$