It's easy to compute the modulus
$$\bigg|z^3|z|(1-i)\frac{\overline{z}}{2}\bigg| \leq 16\sqrt{2} \iff |z| \leq 2$$
But I always have trouble with the argument,
$$\Im\bigg(z^3|z|(1-i)\frac{\overline{z}}{2}\bigg)<0 \iff \Im(z^2(1-i)) < 0$$
If $z=re^{\theta i}$, then $z^2(1-i)=\sqrt{2}r^2e^{2\theta + \frac{3\pi}{4}i}$. So we want to solve
$$\pi < 2\theta + \frac{3\pi}{4} - 2k\pi< 2\pi \iff \\ k\pi - \frac{7}{8}\pi < \theta < \frac{\pi}{8} + k\pi$$
Does this mean that $\theta \in (\frac{\pi}{8},\frac{9\pi}{8}) $? I'm super confused.
You have made a mistake, $$ 1 - i = \sqrt{2} e^{-i \tfrac{\pi}{4}} \text{ or } \sqrt{2} e^{i \tfrac{7\pi}{4}}$$
If I choose to proceed my the former convention, i.e. $ - \pi \le arg(z) \le \pi$, then I am supposed to solve $$ -\pi \le 2 \theta - \cfrac{\pi}{4} \le 0$$
$$ \implies -k \pi - \cfrac{3 \pi}{8} \le \theta \le -k \pi + \cfrac{\pi}{8} $$
Ofcourse even on following the latter convention th eresult would be same.