Prove the following

According to the question,

We have,

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

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

$=\left[\frac{(1-i)^{2}}{1-i^{2}}\right]^{100}$

$=\left(\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1}\right)^{100}$

$=\left(-\frac{2 \mathrm{i}}{2}\right)^{100}$

$=\left(i^{4}\right)^{25}$

= 1

Hence, (a, b) = (1, 0)