If a+bi = frac{i}{1-i},then (a, b) = | vedclass

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(D) Given,$a+bi = \frac{i}{1-i}$ Multiply the numerator and denominator by the conjugate of the denominator $(1+i)$: $a+bi = \frac{i(1+i)}{(1-i)(1+i)}

Vedclass · Accepted Answer

$(\frac{-1}{2}, \frac{1}{2})$

Vedclass · Answer

(D) Given,$a+bi = \frac{i}{1-i}$ Multiply the numerator and denominator by the conjugate of the denominator $(1+i)$: $a+bi = \frac{i(1+i)}{(1-i)(1+i)}$ $a+bi = \frac{i+i^2}{1^2-i^2}$ Since $i^2 = -1$: $a+bi = \frac{i-1}{1-(-1)} = \frac{-1+i}{2}$ $a+bi = \frac{-1}{2} + \frac{1}{2}i$ Comparing the real and imaginary parts,we get $a = \frac{-1}{2}$ and $b = \frac{1}{2}$ Therefore,$(a, b) = (\frac{-1}{2}, \frac{1}{2})$

If $a+bi = \frac{i}{1-i}$,then $(a, b) =$

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