The modulus-amplitude form of frac{(1-i)^3(2- | vedclass

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(A) Let $z = \frac{(1-i)^3(2-i)}{(2+i)(1+i)}$.First,simplify the numerator: $(1-i)^2 = 1 - 2i + i^2 = -2i$. So,$(1-i)^3 = -2i(1-i) = -2i + 2i^2 = -2 -

Vedclass · Accepted Answer

$2 \operatorname{cis}\left(\pi-	an ^{-1} \frac{4}{3}ight)$

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(A) Let $z = \frac{(1-i)^3(2-i)}{(2+i)(1+i)}$.First,simplify the numerator: $(1-i)^2 = 1 - 2i + i^2 = -2i$. So,$(1-i)^3 = -2i(1-i) = -2i + 2i^2 = -2 - 2i$.Then,$(1-i)^3(2-i) = (-2-2i)(2-i) = -4 + 2i - 4i + 2i^2 = -4 - 2i - 2 = -6 - 2i$.Next,simplify the denominator: $(2+i)(1+i) = 2 + 2i + i + i^2 = 2 + 3i - 1 = 1 + 3i$.Now,$z = \frac{-6-2i}{1+3i} = \frac{(-6-2i)(1-3i)}{(1+3i)(1-3i)} = \frac{-6 + 18i - 2i + 6i^2}{1^2 + 3^2} = \frac{-6 + 16i - 6}{10} = \frac{-12 + 16i}{10} = -1.2 + 1.6i$.The modulus $r = \sqrt{(-1.2)^2 + (1.6)^2} = \sqrt{1.44 + 2.56} = \sqrt{4} = 2$.The complex number lies in the second quadrant. The argument $	heta = \pi - 	an^{-1}\left(\frac{1.6}{1.2}ight) = \pi - 	an^{-1}\left(\frac{4}{3}ight)$.Thus,the modulus-amplitude form is $2 \operatorname{cis}\left(\pi - 	an^{-1} \frac{4}{3}ight)$.

The modulus-amplitude form of $\frac{(1-i)^3(2-i)}{(2+i)(1+i)}$ is

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