I am reading Harold Widom's paper "Extremal Polynomials Associated with a System of Curves in the Complex Plane". At the beginning of section 11 he states that:

[There is] a simple transformation which opens up the arc. Suppose $E$ is a Jordan arc which for simplicity we assume has endpoints $\pm 1$. Then if $(z^2-1)^{1/2}$ denotes the branch of the square root in the complement of $E$ which is asimptotically $z$ near $z=\infty$ we set $s=z+\sqrt{z^2-1}$ and $z=\frac{1}{2}(s+s^{-1})$. Then the exterior of $E$ corresponds to the exterior of a certain closed curve $E'$ in the $s$-plane."

I was thinking a long time about this "simple" fact and I cannot convince myself about its certainty. The statement is true with those mappings and $E=[-1,1]$. But, as I understand the mappings works as well for any other arc with the same endpoints.So, I took the upper semi-circumference and I draw its image curve by the map $s=z+\sqrt{z^2-1}$, and the result was an arc, not a contour. Where is the problem then?