Let $\gamma$ be a continuous curve in the complex plane without self-intersections and let $\lambda$ be a complex non-real number less than 1 in modulus. Put $\gamma'=\lambda\gamma$.


**Question.** Is it true that the sumset $\gamma+\gamma'$ has a non-empty interior? 



The Minkowski sum of two curves
Well, I cannot call it a complete answer, but it seems an idea that can work.


First: take any loop in the (s,t) coordinates, and consider its image on the complex plane under $\gamma(s)+\gamma'(t)$. Any point having nonzero index w.r.t. this loop is then contained in $\gamma+\gamma'$.


Second: take the loop to be the boundary of a rectangle, $[s\_1,s\_2]\times [t\_1,t\_2]$, and assume that along the second coordinate it is very thin, $t\_1$ is very close to $t\_2$. Then, the boundary consists of two translated images $\gamma([s\_1,s\_2])+\gamma'(t\_1)$ and $\gamma([s\_1,s\_2])+\gamma'(t\_2)$, and two very small curves $\gamma(s\_j)+\gamma'([t\_1,t\_2])$.


Third: I would say that the two translated images should more or less coincide, up to what happens at the endpoints. As if they do not, there is a part of one "somewhere in the middle", where it is far away from the other one, and then, crossing the curve, we change the index -- so on one of two sides there will be an open set of points with nonzero index.


Last: making $t\_2$ go towards $t\_1$, in the limit of the almost-translation-invariance above we get that $\gamma([s\_1,s\_2])$ is a straight line. And this allows to conclude easily, as $\lambda$ is not real.


Well, as I've said, it's a sketch, but it has a good chance of working.


**Upd.:** I'm now sure that this argument works. For "somewhere in the middle", I mean the following.


**Lemma.** Let $\delta$ be the diameter of $\gamma'([t\_1,t\_2])$, and assume that there is no point with nonzero index w.r.t. the loop described in the "*Second.*" step. Then, the sets $\gamma([s\_1,s\_2])+\gamma'(t\_1)$ and $\gamma([s\_1,s\_2])+\gamma'(t\_2)$ coincide outside $\delta$-neighborhoods of $\gamma'(t\_{j})$.


**Sketch of the proof.** Assume the contrary and let $z$ be a point of a curve $\gamma([s\_1,s\_2])+\gamma'(t\_1)$ that lies outside the above neighborhoods and that does not belong to the $\gamma([s\_1,s\_2])+\gamma'(t\_2)$. Due to the continuity, there is $\epsilon$-neighborhood of $z$ that the latter curve does not intersect. 


Now, in this neighborhood one can find two points "on different sides" with respect to the first curve (yes, this phrase should also be more formally stated with an appropriate reference to the Jordan curve theorem). Such two points have thus different index w.r.t. all the closed loop (image under $\gamma(s)+\gamma'(t)$ of the boundary of the rectangle $[s\_1,s\_2]\times [t\_1,t\_2]$). Hence, at least one of them in an internal point of the sum $\gamma+\gamma'$. $\square$


Passing to the limit as $t\_2\to t\_1$ says that the curve $\gamma([s\_1,s\_2])$ should admit (if there is no internal point) arbitrarily small translational symmetries outside arbitrarily small neighborhoods of its endpoints. Hence, it is a straight segment.


