<s>[INST] Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y^2-ix+y)\right|$
$\left|3+2x^2-2y^2+y\right|$
$3+2x^2-2y^2+y$
And then I'm not really sure where to go. Pretty sure I took the wrong approach because not using the initial information that $|z| \le 1$
 [/INST] You need to prove $\left|3+2x^2-2y^2+y\right|\le 6$ subject to $|z|\le 1 \iff x^2+y^2\le 1$.
Note:
$$\left|3+2x^2-2y^2+y\right|=\left|3+2(x^2+y^2)-4y^2+y\right|\le |5-4y^2+y|=|4y^2-y-5|\le6 \Rightarrow \\
\begin{cases}4y^2-y-5\ge -6\\ 4y^2-y-5\le 6\end{cases} \Rightarrow  \begin{cases}4y^2-y+1\ge 0\\ 4y^2-y-11\le0\end{cases} \Rightarrow \begin{cases}y\in(-\infty,+\infty)\\ y\in\left[\frac{1-\sqrt{177}}{8}\approx -1.538,\frac{1+\sqrt{177}}{8}\approx 1.788\right]\end{cases}.$$
Also note:
$$x^2+y^2\le 1 \Rightarrow -1\le y\le 1.$$
</s>