The conjugate of the complex number frac{(1+i | vedclass

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(D) Let $z = \frac{(1+i)^{2}}{1-i}$.First,expand the numerator: $(1+i)^{2} = 1^{2} + i^{2} + 2i = 1 - 1 + 2i = 2i$.So,$z = \frac{2i}{1-i}$.To simplify

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$-1-i$

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(D) Let $z = \frac{(1+i)^{2}}{1-i}$.First,expand the numerator: $(1+i)^{2} = 1^{2} + i^{2} + 2i = 1 - 1 + 2i = 2i$.So,$z = \frac{2i}{1-i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(1+i)$:$z = \frac{2i(1+i)}{(1-i)(1+i)} = \frac{2i + 2i^{2}}{1 - i^{2}} = \frac{2i - 2}{1 - (-1)} = \frac{2i - 2}{2} = i - 1$.Thus,$z = -1 + i$.The conjugate of a complex number $z = a + bi$ is $\bar{z} = a - bi$.Therefore,the conjugate of $-1 + i$ is $-1 - i$.

The conjugate of the complex number $\frac{(1+i)^{2}}{1-i}$ is

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