Question

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

Answer 1

$$\begin{gathered} \frac{1-i}{1+i}=\frac{1-i}{1+i}\times \frac{1-i}{1-i} \\ =\frac{{\left(1-i\right)}^2}{1^2-i^2} \\ =\frac{1^2+i^2-2i}{1+1}[\because i^2=-1] \\ =\frac{1-1-2i}{2} \\ =\frac{-2i}{2} \\ =-i....\left(1\right) \end{gathered}$$


It is given that,
$$\begin{gathered} {\left(\frac{1-i}{1+i}\right)}^{100}=a+ib \\ \Rightarrow (-i{)}^{100}=a+ib[\mathrm{From}\left(1\right)] \\ \Rightarrow i^{4\times 25}=a+ib \\ \Rightarrow 1+0i=a+ib[\because i^4=1] \\ \Rightarrow a=1\mathrm{and}b=0 \end{gathered}$$

Thus, (a, b) = (1, 0).