October 13, 2006

Find the smallest positive number $A$ such that

\frac{21}{|z^4-5z^2+6|}\le A

for every complex number $z$ on the circle $|z|=3$. Prove that $A$ is the smallest such number.

Solution. We have that

|z^4-5z^2+6|=|(z^2-3)(z^2-2)|=|z^2-3||z^2-2|.

We also know that $|z_1 - z_2|\ge ||z_1|-|z_2||$ for all $z_1$ and $z_2$ in $\mathbb{C}$. Therefore,

|z^2-3|\ge ||z^2|-3|=||z|^2-3|=|9-3|=6

and

|z^2-2|\ge ||z^2|-2|=||z|^2-2|=|9-2|=7.

Thus, $|z^2-3||z^2-2|\ge 6\cdot 7 = 42$, and finally,

\frac{21}{|z^4-5z^2+6|}\le \frac{21}{42}=\frac12.

So $A=1/2$ is an upper bound. To show that this is the smallest such number, we merely note that we achieve equality above if $z=\pm 3$.