Record, (Full marks 20 points) 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.
Let $z-i=x+(y-1) i=r(\cos \theta+i \sin \theta)$, then
$$
|z-i|=r, 0 \leqslant r \leqslant 1 \text {. }
$$

So $x=r \cos \theta$.
Then $A=r \cos \theta\left(r^{2}-1\right)=r\left(r^{2}-1\right) \cdot \cos \theta$,
$$
-r\left(1-r^{2}\right) \leqslant A \leqslant r\left(1-r^{2}\right),
$$

Thus $A^{2} \leqslant \frac{1}{2} \cdot 2 r^{2}\left(1-r^{2}\right)\left(1-r^{2}\right)$
$$
\leqslant \frac{1}{2}\left[\frac{2 r^{2}+\left(1-r^{2}\right)+\left(1-r^{2}\right)}{3}\right]^{3}=\frac{4}{27},
$$

which means $-\frac{2 \sqrt{3}}{9} \leqslant A \leqslant \frac{2 \sqrt{3}}{9}$.
Equality holds when $2 r^{2}=1-r^{2}$, i.e., $r=\frac{\sqrt{3}}{3}$.
When $\cos \theta=-1$, $A>0$; when $\cos \theta=1$, $A<0$.
When $i=-\frac{\sqrt{3}}{3}+i$, $A_{\text {max }}=\frac{2 \sqrt{3}}{9}$.

When $x=\frac{\sqrt{3}}{3}+i$, $A_{\text {min}}=-\frac{2 \sqrt{3}}{9}$.