If {left( {frac{{1 - i}}{{1 + i}}} right)^{10 | vedclass

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(B) Given,${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$. First,simplify the fraction inside the bracket: $\frac{1 - i}{1 + i} = \frac{(1

Vedclass · Accepted Answer

$a = 1, b = 0$

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(B) Given,${\left( {\frac{{1 - i}}{{1 + i}}} ight)^{100}} = a + ib$. First,simplify the fraction 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 expression: $(-i)^{100} = (-1)^{100} 	imes i^{100} = 1 	imes (i^4)^{25} = 1 	imes (1)^{25} = 1$. Thus,$a + ib = 1 + 0i$. Comparing the real and imaginary parts,we get $a = 1$ and $b = 0$.

If ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$,then

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