The imaginary part of frac{(1-i)^3}{(2-i)(3-2 | vedclass

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Question

(D) Let $Z = \frac{(1-i)^3}{(2-i)(3-2i)}$.First,expand the numerator: $(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.Next,ex

Vedclass · Accepted Answer

$-\frac{22}{65}$

Vedclass · Answer

(D) Let $Z = \frac{(1-i)^3}{(2-i)(3-2i)}$.First,expand the numerator: $(1-i)^3 = 1^3 - 3(1)^2(i) + 3(1)(i)^2 - i^3 = 1 - 3i - 3 + i = -2 - 2i$.Next,expand the denominator: $(2-i)(3-2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i - 2 = 4 - 7i$.So,$Z = \frac{-2 - 2i}{4 - 7i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator: $Z = \frac{-2 - 2i}{4 - 7i} 	imes \frac{4 + 7i}{4 + 7i} = \frac{-8 - 14i - 8i - 14i^2}{16 + 49} = \frac{-8 - 22i + 14}{65} = \frac{6 - 22i}{65}$.Thus,$Z = \frac{6}{65} - \frac{22}{65}i$.The imaginary part is $-\frac{22}{65}$.

The imaginary part of $\frac{(1-i)^3}{(2-i)(3-2i)}$ is

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