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

$\frac{{(1-i)}^3(2-i)}{(2+i)(1+i)}=\frac{{(1-i)}^4{(2-i)}^2}{(4+1)(1+1)}=\frac{1}{10}{(1+i^2-2i)}^2+(4+i^2-4i)=\frac{1}{10}{(-2i)}^2(3-4i)$

$=-\frac{4}{10}(3-4i)=2\left(-\frac{{\displaystyle 3}}{{\displaystyle 5}}+\frac{{\displaystyle 4}}{{\displaystyle 5}}i\right)=2(\cos \theta +i\sin \theta )$ where $\cos \theta =-\frac{3}{5},\sin \theta =\frac{4}{5}$

$\therefore$The amplitude $\theta$ lies in 2nd quadrant and $\tan \theta =\frac{4}{3}$, so that $\theta =\pi -{\tan }^{-1}\left(\frac{{\displaystyle 4}}{{\displaystyle 3}}\right)$

$\therefore z=2cis\left[\pi -{\tan }^{-1}(4/3)\right]$