Let $z$ be a complex number satisfying $|z|^4-16|z|^2-3(z^2+{\overline{z}}^2)+9=0$ .. 
Let $z$ be a complex number satisfying $$|z|^4-16|z|^2-3(z^2+{\bar {z}}^2)+9=0$$ and let $\min(|z|)=a,b=\max(|z|)$ then find $ab$.


We notice that $$\Im(z^2+{\overline{z}}^2)=0$$
$$\Im({(z+\overline{z})}^2-2z\bar z)=0$$ but this is obvious as both $z+\bar z,z \bar z$ are real.
Hence i was unable to find any additional condition
Also substituition $z=x+iy$ and taking $x^2=l,y^2=m$ we have the mess $$l^2+m^2-22l+10m+2lm+9=0$$I recognise this as a quadratic and setting $D\ge 0$  may give the max and min value of $l+m$.
But surely there must be something easier?!

First let $t = z^2$, so that we can rewrite that as
$$|t|^2-16|t| - 3(t+\overline t) + 9 = 0 \iff (|t| - 8)^2 = 3(t + \overline t) + 55$$
Now it may be easier to note that $|t|$ can find extrema when $\Im(t)=0$, so that we may consider only the case $t \in \mathbb R$.  So we have for extreme $|t| = k$, the quadratic:
$$k^2 -22k +9 = 0 \implies |t|_\min \cdot |t|_\max = 9$$
Hence $a\cdot b = \sqrt{9}=3$
