A function $f:\mathbb{R}^2\to \mathbb{R}$ has the property $(\mathcal{P})$ if:
       For any finitely many points $A_1,A_2, \dots, A_n$ such that $f(A_1)=f(A_2)=\dots=f(A_n)$, it        follows that $A_1,A_2, \dots ,A_n$ form a convex polygon.
Let $Q\in \mathbb{C}[X]$. Prove that the function $f(x,y)=|Q(x+iy)|$ has the propery $(\mathcal{P})$ if and only if all roots of $Q$ are equal.

I tried to think of this geometrically in 3 dimensions, but I honestly don't even know how to approach it. If we take $n$ complex numbers wich do not form a convex polygon, then we are assured that not all $|Q(z_1)|,\dots ,|Q(z_n)|$ are equal. I don't know how to proceed.

  • $\begingroup$ The definition is confusing for me: does it mean that there are finite point $\;A_1,...,A_n\;$ s.t. ...etc., or for any finite number of points s.t ....etc.? $\endgroup$ – DonAntonio Feb 14 '18 at 13:34
  • $\begingroup$ "For any"! I'll edit the question. $\endgroup$ – AndrewC Feb 14 '18 at 13:40
  • $\begingroup$ Hint: consider the set $\{z:|Q(z)|=c\}$. What would happen for small $c$ if $Q$ had distinct roots? $\endgroup$ – Wojowu Feb 14 '18 at 13:45
  • $\begingroup$ This is a problem in geometric functional equation.A similar problem can be found in Romania TST 1996 paper. $\endgroup$ – Shubhrajit Bhattachrya Feb 14 '18 at 13:46

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