Can this simple integral be zero for a Jordan curve?

Question

The following simple problem came up while doing some unrelated research.

Does there exist a Jordan curve $\gamma : [0,2\pi] \to \mathbb{C}$ of positive orientation, lets say $C^1$-smooth (just to simplify the problem) that satisfies:

$$\int_0^{2\pi} \gamma(t)e^{-it} dt = 0 $$

I am hoping that such a curve does not exist, but I could be overlooking something simple.

Answer 1

Yes it exists.

Think of the curve which goes from $1$ to $2+\varepsilon{\cdot} i$ and then to $-1$ and which is symmetric with respect to real axis; so $\gamma(-t)=\bar \gamma(t)$ and it is defined in $[-\pi,\pi]$.

For such curves the integral is real.

• If we run the arc from $2- \varepsilon{\cdot} i$ to $2+\varepsilon{\cdot} i$ too fast then your integral is positive.
• if we run it slow and spend a lot of time near $\pm\pi$ close to $2$ then your integral is negative.

So somewhere you will get zero.