2
$\begingroup$

Let $P_1$ and $P_2$ be complex polynomials with complex coefficients and $c > 0$. Can we find polynomial $P_3$ and $c’>0$ such that

$\{z \in \mathbb C : |P_1(z)| \geq c\} \cap \{ z \in \mathbb C : |P_2(z)| \geq c\}= \{ z \in \mathbb C : |P_3(z)| \geq c’\}$

holds?

It seems like this is either trivially false or very hard problem. I don’t have much idea. Any suggestion or reference is welcome. Thanks!

$\endgroup$
3
$\begingroup$

You are right: this is trivially false.

Let $P_1(z)=z^2,\; P_2(z)=(z-a)^2$, and $c=1$. Then the boundaries of the first two level sets are circles of radius $1$, and choosing an appropriate $a$ you can make them cross at any given angle.

On the other hand the boundary of the set in the RHS is a polynomial lemniscate, and it is clear that it cannot cross itself under any angle except $\pi/n$ where $n$ is an integer. (This is a general local property of level sets of harmonic functions, which is easy to prove.)

$\endgroup$
2
  • $\begingroup$ I didn’t understand your comment in the bracket because $|P_3|$ need not be harmonic. Also can we at least find polynomial $P_3$ whose superlevel set is close (may be in Housdorff metric sense) to intersection of superlevel sets of $P_1$ and $P_2$? $\endgroup$
    – Mayuresh L
    Apr 19 '19 at 11:50
  • 1
    $\begingroup$ @Mayuresh: $\log|P|$ is harmonic (away from the zeros of $P$) and has the same level lines. $\endgroup$ Apr 19 '19 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.