Question:
If $\left(\frac{1-i}{1+i}\right)^{100}=a+i b$, find $(a, b)$
Solution:
$\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$
$=\frac{(1-i)^{2}}{1^{2}-i^{2}}$
$=\frac{1^{2}+i^{2}-2 i}{1+1}$ $\left[\because i^{2}=-1\right]$
$=\frac{1-1-2 i}{2}$
$=\frac{-2 i}{2}$
$=-i$ ....(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, $(a, b)=(1,0)$.
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