Solve this

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

Consider the given equation,

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

Now, we rationalize

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

[Here, we multiply and divide by the conjugate of 1 + i]

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

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

Using $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

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

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

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

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

$=(-i)^{100}$

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

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

$=(1)^{25}$

$\left[\because i^{4}=j^{2} \times i^{2}=-1 \times-1=1\right]$

$(a+i b)=1+0 i$

On comparing both the sides, we get

a = 1 and b = 0

Hence, the value of a is 1 and b is 0