If left(frac{1-i}{1+i}right)^{96}=a+ib,then ( | vedclass

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(B) Given that,$\left(\frac{1-i}{1+i}\right)^{96}=a+ib$.First,simplify the expression inside the bracket:$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-

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$(1, 0)$

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(B) Given that,$\left(\frac{1-i}{1+i}\right)^{96}=a+ib$.First,simplify the expression inside the bracket:$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$.Now,substitute this back into the equation:$(-i)^{96} = a+ib$.Since $96$ is an even number,$(-i)^{96} = i^{96}$.$i^{96} = (i^4)^{24} = (1)^{24} = 1$.So,$1 = a+ib$,which can be written as $1+0i = a+ib$.Comparing the real and imaginary parts,we get $a=1$ and $b=0$.Therefore,$(a, b) = (1, 0)$.

If $\left(\frac{1-i}{1+i}\right)^{96}=a+ib$,then $(a, b)$ is

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