Show that if $|z|=2$, $\text{Im}(1-\bar{z}+z^2)\le 7$. Show that if $|z|=2$, $|\text{Im}(1-\bar{z}+z^2)|\le 7$.
My attempt:
Let $z=x+yi$, $|z|=\sqrt{x^2+y^2}$, $|z|^2=(\sqrt{x^2+y^2})^2=x^2+y^2$.
Since $|z|=2$, then $|z|^2=2^2=4$. So $x^2+y^2=4$. Also, $(x-y)^2\ge0$ $\implies x^2+y^2\ge2xy$;
$$xy\le2 $$
So $|y|\le2$ and $|x|\le 2$.
Now,
$$\begin{aligned}
1-\bar{z}+z^2&=1-(x-yi)+(x+yi)^2\\
&=1-x+yi+x^2+2xyi-y^2\\
&=1-x+x^2-y^2+(2xy+y)i
\end{aligned}$$
Thus,
$$\begin{aligned}
\ |\text{Im}(1-\bar{z}+z^2)|&=|y+2xy|\\
&\le|y|+|2xy|\\
&\le |y|+|x^2+y^2|\\
&\le2+4=6
\end{aligned}$$
How about the case when it is equal to $7$?

By the triangle inequality:
$$|1 + (-\bar{z}) + z^2 | ≤ |1| + |-\bar{z}| + |z^2| ≤ 1+2+2^2 = 7$$
(think about what happens to $z = 2e^{i \theta}$).
