If z = frac{(2-i)(1+i)^3}{(1-i)^2},then opera | vedclass

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(A) Given $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$.First,simplify $(1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.Next,simplify $(1-i)^2 = 1 - 2i

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$	an^{-1}\left(\frac{1}{3}ight) - \pi$

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(A) Given $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$.First,simplify $(1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.Next,simplify $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.Substituting these into $z$,we get $z = \frac{(2-i)(-2+2i)}{-2i} = \frac{(2-i) \cdot 2(-1+i)}{-2i} = \frac{(2-i)(1-i)}{i}$.Expanding the numerator: $(2-i)(1-i) = 2 - 2i - i + i^2 = 2 - 3i - 1 = 1 - 3i$.So,$z = \frac{1-3i}{i} = \frac{1-3i}{i} \cdot \frac{-i}{-i} = \frac{-i + 3i^2}{1} = -3 - i$.Since $z = -3 - i$ lies in the $3^{	ext{rd}}$ quadrant,the argument is $\operatorname{Arg}(z) = -\pi + 	an^{-1}\left(\frac{-1}{-3}ight) = -\pi + 	an^{-1}\left(\frac{1}{3}ight)$.

If $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$,then $\operatorname{Arg}(z) = $

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