Does there exist a simple smooth closed curve $\gamma:S^1\to \mathbb C$ such that
$$ \int\_{0}^{2\pi} e^{\gamma(e^{it})} \, |\gamma'(e^{it} )|\,dt =0?$$



On a property of complex exponentials
After the change of the variable $w=\gamma(t)$ your integral
becomes
$$\int\_\gamma e^wds,$$
where $ds$ is the length element, $ds=\sqrt{dx^2+dy^2}$, and $\gamma$ is a Jordan curve of zero index about the origin. Now notice that on every vertical segment of length $2\pi k$, where $k$ is an integer,
this integral is zero. Now construct a simple closed curve, of zero index about the origin, consisting of 8 straight segments, 4 vertical and 4 horizontal. All vertical segments have length $2\pi k\_j$
and all hosizontal segments lie on the lines $y=\pi/2+\pi k\_j$, and the curve is symmetric with respect to complex conjugation. Such a curve is easy to draw, and the integral along it is 0. Indeed it is zero on each vertical segment,
while contributions from horizontal segments cancel.


Let me add how to make the curve smooth. We will slightly modify the original curve, keeping the vertical segments unchanged, so their contribution to
the integral is zero. We will also keep the symmetry
with respect to the complex conjugation, so that imaginary part of the integral will be 0.


First we choose the vertical segments $\mathrm{Re} z=x\_k$ so that for some segments $\cos x\_j<0$ and for
some other $\cos x\_j>0$, so that the contribution to
the integral of some segments is positive and for other segments it is negative. Then, to make the curve smooth, we replace some small pieces of horizontal segments near the corners by pieces which make the curve smooth. Since the contribution of some of these pieces is negative and others positive, it is clear that the pieces can be chosen so that their contribution cancels.


