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 $|z| \le 1$

Note that $3+2x^2-2y^2+y\leq5+y-4y^2$ if $x^2+y^2\leq 1$.  Therefore,
$$3+2x^2-2y^2+y\leq 5+y-4y^2=5+\frac{4y(1-4y)}{4}\leq 5+\frac{1}{4}\,\left(\frac{4y+(1-4y)}{2}\right)^2=\frac{81}{16}$$
for $y\in[-1,+1]$, where we have applied the AM-GM Inequality on the right-hand inequality.  We also have $3+2x^2-2y^2+y\geq 3+y-2y^2$, so
$$3+2x^2-2y^2+y\geq 3+y-2y^2=(3-2y)(1+y)\geq 0\text{ for all }y\in[-1,+1]$$  This shows that
$$0\leq \text{Re}\big(3+\text{i}+2\,\bar{z}^2-\text{i}\,z\big)\leq \frac{81}{16}$$
for all complex numbers $z$ such that $|z|\leq 1$.  The left-hand side is an equality iff $z=-\text{i}$, and the right-hand side is an equality iff $z=\dfrac{\pm3\sqrt{7}+\text{i}}{8}$
