If

$\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$=\frac{(1-i)^{2}}{1^{2}-i^{2}}$

$=\frac{1^{2}+i^{2}-2 i}{1+1}$    $\left[\because i^{2}=-1\right]$

$=\frac{1-1-2 i}{2}$

$=\frac{-2 i}{2}$

$=-i$    ....(1)

It is given that,

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

$\Rightarrow(-i)^{100}=a+i b \quad[$ From $(1)]$

$\Rightarrow i^{4 \times 25}=a+i b$

$\Rightarrow 1+0 i=a+i b$      $\left[\because i^{4}=1\right]$

$\Rightarrow a=1$ and $b=0$

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