If left(frac{1-i}{1+i}right)^{100}=a+ib,where | vedclass

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(A) First,simplify the expression inside the parenthesis: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+

Vedclass · Accepted Answer

$(1, 0)$

Vedclass · Answer

(A) First,simplify the expression inside the parenthesis: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$. Now,raise this to the power of $100$: $(-i)^{100} = (-1)^{100} 	imes i^{100} = 1 	imes (i^4)^{25} = 1 	imes (1)^{25} = 1$. Given that $a+ib = 1$,we can write this as $1+0i$. Comparing the real and imaginary parts,we get $a=1$ and $b=0$. Thus,$(a, b) = (1, 0)$.

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$,where $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then $(a, b)$ is equal to

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