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

Imaginary part of $\frac{(1-i)^3}{(2-i)(3-2 i)}$ is
  1. $\frac{22}{65}$
  2. $\frac{6}{65}$
  3. $-\frac{6}{65}$
  4. $-\frac{22}{65}$

Solution

$Z=\frac{(1-i)^3}{(2-i)(3-2 i)}=\frac{(1-i)^3}{4-7 i} \times \frac{4+7 i}{4+7 i}$
$\Rightarrow Z=\frac{6-22 i}{65} \Rightarrow \operatorname{Img}(Z)=\frac{-22}{65}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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