0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

I think this an issue of choice of branch of square root. Note that the paper asks you to use the branch holomorphic in $\mathbb{C} \setminus E$ and behaves as $z$ as $z \to \infty$.

Let's call this choice of branch $w(z)$ and call the compact set delimited by $[-1, 1] \cup E$ by $K$. Then the map we are looking for can be defined by $$ z + w(z) = \left\{ \begin{array}{ll} z + \sqrt{z^2 - 1} & \text{ for } z \in \mathbb{C} \setminus K\\ z - \sqrt{z^2 - 1} & \text{ for } z \in \text{Int}(K) \end{array} \right.$$

where $\sqrt{z^2 - 1}$ is the branch holomorphic in $\mathbb{C} \setminus [-1, 1]$. Then, $z + w(z)$ is holomorphic in the complement of $E$ and has the correct behavior as $\infty$. Furthermore, the image of $\text{Int}(K)$ will be some open set in the unit disk and the image of the rest of the plane is in the complement of the unit disk as before.

Roughly speaking, I'm analytically continuing the map $z + \sqrt{z^2 - 1}$ across $[-1, 1]$ to get the correct branch cut.

$\endgroup$
1
  • $\begingroup$ Thanks!! Everything is clear now. $\endgroup$ Apr 9 '20 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.