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To prove $-\frac{1}{2}(z+\frac{1}{z})$ maps upper half disk onto upper half plane, I have been trying to find a formula for the inverse. To this end, I chose $\omega$ in the upper half plane.

$$\omega = -\frac{1}{2}(z+\frac{1}{z})$$ $$0 = z^2+2 \omega z +1$$

At this point I would prefer to use the quadratic formula because I want a formula for the inverse, but I was unable to prove that one of the two solutions given lie inside the unit disk. I have come up with an unsatisfactory existence proof below.

There are two solutions for $z$, call them $z_1$ and $z_2$. By the form of the polynomial $z_1 z_2 = 1$ and $z_1 + z_2 = -2 \omega$. wlog let $|z_1| \lt 1$ and $|z_2| \gt 1$. Since $z_1$ and $z_2$ lie on opposite sides of the real line (because they are inverses), and their sum is in the lower half plane, the smaller one $z_1$ must lie in the upper half plane and hence the upper half disk. So there exists an inverse, but I don't have a closed form solution.

(This is exercise 8.5 in Stein and Shakarchi Complex Analysis )

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  • $\begingroup$ Why do you find your proof unsatisfactory? $\endgroup$ – angryavian Jun 10 '14 at 3:42
  • $\begingroup$ I'd like to have the formula for the inverse. $\endgroup$ – Mark Jun 10 '14 at 3:49
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With $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$:

$$-\frac12\left(z+\frac1z\right)=-\frac12\frac{z^2+1}z\cdot\frac{\overline z}{\overline z}=-\frac12\frac{z|z|^2+\overline z}{|z|^2}=$$

$$=-\frac1{2|z|^2}\left(x(|z|^2+1)+y(|z|^2-1)i\right)$$

The imaginary part of the above is

$$\frac{1-|z|^2}{2|z|^2}y>0\iff |z|<1\;,\;\text{whenever}\;\;y>0\;\ldots\;\text{and we're done.}$$

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  • $\begingroup$ But did you show that every point in the UHP can be reached? And how about a formula for the inverse? Thanks $\endgroup$ – Mark Jun 10 '14 at 3:53
  • $\begingroup$ Why would I want any of those to things, @Mark? I showed that when $\;|z|<1\,,\,\,y>0\;$ (i.e., the upper half unit disk), we map that to the upper half plane, which is what was asked. $\endgroup$ – DonAntonio Jun 10 '14 at 3:55
  • $\begingroup$ Sorry, I meant that I wanted to show the mapping was onto. To do this, I wanted to use the quadratic formula but was unable to make this approach work. I edited the question to clarify. $\endgroup$ – Mark Jun 10 '14 at 4:03
  • $\begingroup$ I realized this is not one my best phrased questions. I edited again to clarify that I want a closed form solution for the inverse. $\endgroup$ – Mark Jan 2 '15 at 2:31
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It turns out Ahlfors has solved this question in his Complex Analysis textbook. I repeat the argument here.

Define $\sqrt{z}$ to be the unique complex number $\zeta$ such that $\Re{\zeta} \gt 0$, defined on $\mathbb{C} - [0, -\infty)$

I will show $z(\omega) = -(\omega - \sqrt{\omega^2 - 1})$ lies inside the unit disk for all $\omega \in \mathbb{C} - [-1, 1]$ and hence provides a closed form expression for the inverse function I was looking for.

This would suffice because we see that if $z(\omega)$ lies in the unit disk like I claim, it must be the smaller of the two roots of the equation $z^2 + 2\omega z + 1 = 0$ (since the other root lies outside of disk by reciprocity), and I already showed that the smaller one would be inside the upper half disk when $\omega$ is in the upper half plane.

Consider the domain $\mathbb{C} - [-1, 1]$. Under the map $$z' = \frac{z+1}{z-1}$$ the domain gets mapped conformally to $\mathbb{C} - [0, -\infty)$ . Then we are able to take the square root with the branch defined previously $$z'' \rightarrow \sqrt{z'}$$ we obtain a map in the plane $\{ \Re z'' \gt 0 \}$ Then finally we map this region conformally to the disk by $$z''' = \frac{z''-1}{z''+1}$$.

In total we get a bijection between ${\sqrt{\frac{z+1}{z-1}} - 1}$ and the unit disk defined $$z'''= \frac{\sqrt{\frac{z+1}{z-1}} - 1}{\sqrt{\frac{z+1}{z-1}} + 1}$$ $$z''' = z - \sqrt{z^2 - 1}$$

And then negating a point in the disk leaves it in the disk so we get what we wanted.

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    $\begingroup$ But how can you show that this maps upper half circle to upper half plane. You only show that this maps $\mathbb{C} \setminus [-1,1]$ to $\mathbb{D}$ $\endgroup$ – mnmn1993 Mar 6 '18 at 17:56
  • $\begingroup$ @mnmn1993 $z(\omega)$ given here satisfies the relation $0 = z^2 + z (\omega) \omega + 1$ and hence the proof in my original post (starting with "There are two solutions...") applies. Namely that $z(\omega)$ is smaller of the two roots of $z^2 + z \omega + 1$ and therefore it lies in the upper half plane for $\omega$ in the upper half plane. $\endgroup$ – Mark Mar 10 '18 at 23:52
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Ignoring the $-\frac{1}{2}$ you have there, you basically want to find the inverse of the Joukowski map $\phi (z)= z+\frac{1}{z}$.

Look at this question (and its answer):

Inverse of the Joukowski map $\phi(z) = z + \frac{1}{z}$

You may also be interested in this paper for more information:

http://www.jpier.org/PIERL/pierl19/13.10091305.pdf

[2010. INVERSE JOUKOWSKI MAPPING. C.-H. Liang, X.-W. Wang, and X. Chen. Science and Technology on Antennas and Microwave Laboratory]

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