Convexity of a set related to certain class of Laurent polynomials 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?
 A: 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:

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 
$\sqrt {245+70\,\sqrt {21}}/35$.
