We know the Joukowski map $$\phi(z) = z + \frac{1}{z}$$ which maps the upper semidisc of radius $1$ in the lower half plane, and the lower semidisc of radius $1$ in the upper half plane.

What is the inverse of this function ? We obtain $z ^{2}-zy + 1 = 0$ and this equation has $2$ solutions, which is the right one ?

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    $\begingroup$ I would think whichever piece sends $-\dfrac{3}{2}i$ to $\dfrac{1}{2}i$ would be for the lower-half plane. That is $\frac{1}{2}(z+\sqrt{z^2-4})$. The other solution for the upper-half. $\endgroup$ – David Peterson Feb 18 '14 at 19:55

We obtain $z^2−zy+1=0$ and this equation has $2$ solutions, which is the right one?

The solutions are given by

$$z = \frac{y - \sqrt{y^2-4}}{2},$$

where the different solutions correspond to the different choices of the square root.

We want a holomorphic inverse, so we need a holomorphic branch of $\sqrt{y^2-4}$ on the image of $\phi$, and the choice of the square root at one point determines whether we land inside the unit circle or outside. Since $\phi(\frac12) = \frac52$, we want

$$1 = \frac{5}{2} - \sqrt{\left(\frac{5}{2}\right)^2-4} = \frac{5}{2} - \sqrt{\frac{25-16}{4}},$$

so we need the branch of $\sqrt{y^2-4}$ on $\mathbb{C}\setminus [-2,2]$ that takes the value $\frac{3}{2}$ in $y = \frac{5}{2}$.

  • $\begingroup$ Aren't there infinitely many branches of $\sqrt {} $? We can take our cut any which way we like. Maybe you mean that the only branches to consider are the principal branch and its negative, however it wasn't obvious to me that one of these would work. But today I saw a proof which I copied in my answer here math.stackexchange.com/questions/828711/… $\endgroup$ – Mark Jan 2 '15 at 2:32
  • $\begingroup$ @Mark On every connected open set, there are exactly two branches of the square root of a function (if any). Of course you have infinitely many choices for the branch cut, but once you have chosen the cut, only the value at one point is free to be chosen. In the situation here, the natural choice of the branch cut is $[-2,2]$, then with that choice of the cut, we have two branches to choose from. $\endgroup$ – Daniel Fischer Jan 6 '15 at 17:18

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