Let $A = (2, 0)$, $B = (0, 2)$, $C = (-2, 0)$, and $D = (0, -2)$.  Compute the greatest possible value of the product $PA \cdot PB \cdot PC \cdot PD$, where $P$ is a point on the circle $x^2 + y^2 = 9$.
We use complex numbers. Let $a = 2$, $b = 2i$, $c = -2$, and $d = -2$ be the complex numbers corresponding to $A$, $B$, $C$, and $D$, respectively. Let $p$ be the complex number corresponding to $P$, so that $|p| = \sqrt{9} = 3$. Then we have \[\begin{aligned} PA \cdot PB \cdot PC \cdot PD &= |p-2| \cdot |p-2i| \cdot |p+2| \cdot |p+2i| \\ &= |(p-2)(p+2)| \cdot |(p-2i)(p+2i)| \\ &= |p^2-4| \cdot |p^2+4| \\ &= |p^4-16|. \end{aligned}\]Since $|p| = 3$, we have $|p^4| = 3^4= 81$, so by the triangle inequality, \[|p^4-16| \le |p^4| + |-16| = 81 + 16 = 97.\]Equality holds if and only if $p^4 = -81$, which occurs when $p = 3\left(\frac{\sqrt2}{2} + \frac{\sqrt2}{2}i\right)$. Therefore, the answer is $\boxed{97}$.