Question

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

Answer 1

$\frac{1-i}{1+i}$
$=\frac{1-i}{1+i}\times \frac{1-i}{1-i}$
$=\frac{1+i^2-2i}{1^2-i^2}$
$=-i$
$\therefore \frac{1-i}{1+i}=-i...\left(1\right)$
$\because {\left(\frac{1-i}{1+i}\right)}^{100}=a+ib$
$\Rightarrow (-i{)}^{100}=a+ibfromeqn\left(1\right)$
$\Rightarrow 1=a+ib$
$\Rightarrow a=1,b=0$
{equating the real and imaginary parts}
Final answer: (1,0)