6 - 123 Let the complex number $z=x+i y(x, y \in R)$ satisfy $|z-i| \leqslant 1$. Find the maximum and minimum values of $A$ $=x\left(|z-i|^{2}-1\right)$ and the corresponding $z$ values.
[Solution] Let $z-i=x+(y-1) i=r(\cos \theta+i \sin \theta)$.
Then $\left\{\begin{array}{l}|z-i|=r \leqslant 1,0 \leqslant r \leqslant 1, \\ x=r \cos \theta\end{array}\right.$
Thus,
$$
\begin{array}{l} 
A=r \cos \theta\left(r^{2}-1\right)=r\left(r^{2}-1\right) \cos \theta, \\
A^{2}=r^{2}\left(r^{2}-1\right)^{2} \cdot \cos ^{2} \theta, \\
A^{2} \leqslant r^{2}\left(r^{2}-1\right)^{2} \\
=\frac{1}{2} \cdot 2 r^{2} \cdot\left(1-r^{2}\right)\left(1-r^{2}\right) \\
\leqslant \frac{1}{2}\left[\frac{2 r^{2}+\left(1-r^{2}\right)+\left(1-r^{2}\right)}{3}\right]^{3} \\
=\frac{4}{27}, \\
-\frac{2 \sqrt{3}}{9} \leqslant A \leqslant \frac{2 \sqrt{3}}{9}
\end{array}
$$

Equality holds if and only if $\cos \theta= \pm 1$ and $2 r^{2}=1-r^{2}$.
That is, when $\cos \theta=-1, r=\frac{\sqrt{3}}{3}$, $A=\frac{2 \sqrt{3}}{9}$;
$$
\cos \theta=1, r=\frac{\sqrt{3}}{3} \text {, then } A=-\frac{2 \sqrt{3}}{9} \text {. }
$$

In other words, when $z=-\frac{\sqrt{3}}{3}+i$, $A$ reaches its maximum value $\frac{2 \sqrt{3}}{9}$;
when $z=\frac{\sqrt{3}}{3}+i$, $A$ reaches its minimum value $-\frac{2 \sqrt{3}}{9}$.