For $r,s\in\mathbb{N}$, let $$L(z):=\sum_{j=-r}^{s}a_{j}z^{j}$$ be a Laurent polynomial with real coefficients such that there exists a closed curve $\gamma$ encircling the origin, i.e., $0\in\mbox{Int }\gamma$ (interior of $\gamma$), and $L$ is real valued if restricted to $\gamma$.

Although it is not obvious, there are many of such Laurent polynomials. The most simple example is $L(z)=1/z+z$. Then $\gamma$ is just the unit circle. More involved example is, for instance, $L(z)=2/z+6z+z^{2}$. However, to show the existence of the curve in this case is a more difficult task (but doable).

My conjecture is that, if $L$ has the property as above, then $$\mbox{Int }\gamma \textbf{ is a convex set}.$$ Can you prove/disprove it?

  • $\begingroup$ Why would you expect this to be true? $\endgroup$ – Anthony Quas Sep 22 '16 at 20:40
  • $\begingroup$ This is based mainly on a quite convincing numerical evidence and partly on my own intuition. $\endgroup$ – Twi Sep 22 '16 at 20:45
  • $\begingroup$ What you're asking about is basically the set of points $L^{-1}(\mathbb R)$ where $L$ takes real values. The interesting points are the points on this curve where $L'(z)=0$ - these are points where two or more different curves cross. The interesting cases to test would be where there are lots of points on $L^{-1}(\mathbb R)$ with $L'(z)=0$; maybe also higher order 0's. How complicated are your examples? $\endgroup$ – Anthony Quas Sep 22 '16 at 21:34
  • $\begingroup$ Would someone be so kind as to define an encircling curve? $\endgroup$ – Włodzimierz Holsztyński Sep 23 '16 at 2:03
  • $\begingroup$ 2Anthony Quas: Yes, you're right as it has been show by Robert Israel below on a concrete example. The test examples which I have considered were of high degree ($r+s$) but they lack critical points of higher degree, indeed. $\endgroup$ – Twi Sep 26 '16 at 18:09

Consider $$ L = (z^2+1)^3 (q^2 z^2 -1)/z$$ With $z = x+iy$, $\text{Im}(L)=0$ on the rather nasty-looking curve $Q(x,y)=0$, where $$ Q(x,y) = 1+7\,{q}^{2}{x}^{8}-28\,{q}^{2}{x}^{6}{y}^{2}+ \left( 15\,{q}^{2}-5 \right) {x}^{6}-14\,{q}^{2}{x}^{4}{y}^{4}+ \left( -15\,{q}^{2}+5 \right) {x}^{4}{y}^{2}+ \left( 9\,{q}^{2}-9 \right) {x}^{4}+20\,{q}^{ 2}{x}^{2}{y}^{6}+ \left( -27\,{q}^{2}+9 \right) {x}^{2}{y}^{4}+ \left( 6\,{q}^{2}-6 \right) {x}^{2}{y}^{2}+ \left( {q}^{2}-3 \right) {x}^{2}-{q}^{2}{y}^{8}+ \left( 3\,{q}^{2}-1 \right) {y}^{6}+ \left( -3 \,{q}^{2}+3 \right) {y}^{4}+ \left( {q}^{2}-3 \right) {y}^{2} $$ Take $q = 2 \sqrt{3}-\sqrt{7}$. This is chosen so that $Q(x,0)$ has roots of order $2$ at $x =\pm\sqrt {245+70\,\sqrt {21}}/35$, while $Q(0,y)$ has roots of order $3$ at $y=\pm 1$. The result is that the graph of $Q(x,y)=0$ looks like this:

enter image description here

There is a closed curve enclosing the origin, but it is (slightly) non-convex. Note that near $x=0$, $y=1$ we have $Q(x,y) \sim 8 (q^2+1)(3 x^2 (y-1) - (y-1)^3)$, so the curve comes in to $(0,1)$ from the lower right tangent to $x = (1-y)/\sqrt{3}$. This tangent hits $y=0$ at $x = 1/\sqrt{3}$, which is to the left of the intersection of the curve with the $x$ axis at $\sqrt {245+70\,\sqrt {21}}/35$.

  • $\begingroup$ This is a very nice counter-example, thank you Robert Israel! $\endgroup$ – Twi Sep 26 '16 at 18:03

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