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

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


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!


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.)

  • $\begingroup$ 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

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