# Conformal Mappings from $xy=1$ to the upper half plane?

For a complex analysis class, I need to find a bijective conformal mapping that maps the region between the hyperbolas $xy = 1$ to the upper half plane.

Any ideas? I'm trying to map the curve to $y=0$ as a first step... but no such luck... reference texts and internet queries only turn up discs, rectangular regions, wedges, and the like.

Much appreciated.

Hint: consider $f(z) = z^2.$ What are the real and imaginary parts?

• Are you suggesting you can only solve this problem if you happened to recall a bit of trivia about $z^2$? – user14972 Dec 11 '13 at 18:50
• @Hurkyl OK, map the region above the sinusoid $y=\sin x$ to the unit disk. Closed form, please. – Igor Rivin Dec 11 '13 at 18:53
• @Hurkyl ps: $z^2$ is a basic function (in complex analysis, but in other fields of human endeavor, too), so I would not be so rash as to call knowledge of the real and imaginary parts thereof "trivia". – Igor Rivin Dec 11 '13 at 18:55
• I think that if you happen to have studied complex analysis out of Ahlfors, you will be familiar with the mapping properties of basic functions. Certainly the power functions are among these. – Lubin Dec 11 '13 at 21:44

Even with @IgorRivin’s helpful hint, this is not so easy.

What his hint tells us to do is square $z$. Then the original region, between the two branches of the hyperbola $xy=1$, maps onto the half-plane $y<2$. But two-to-one! This is something of a problem, and it took me a while to get around it.

Instead, perform the operation $z\mapsto4/z^2$. This sends our region to the exterior of the circle with center $-i$, radius $1$. This sends $\pm(1+i)$ to $-2i$ and $+\infty$ (limit of positive reals, getting bigger and bigger) to $0^+$, while $+\infty i$ (limit as you go up the imaginary axis) to $0^-$, where these two strange notations mean that you get towards zero from the right and from the left along the real axis, respectively.

So we’ve mapped, still two-to-one, onto the exterior of this circle. The advantage is that the ramification point now is at $\infty$, and we can “deramify” in the following way: add $i$ to our values, now $z\mapsto i+4/z^2$, double cover of the exterior of the unit circle, ramifying at infinity. Now just take the square root, that will presumably give an unramified mapping of our region onto the exterior of the unit circle. Finally take reciprocal and simplify: $$z\mapsto\frac1{\sqrt{i+\frac4{z^2}}}=\frac z{\sqrt{4+iz^2}}$$ This maps the region one-to-one onto the unit disk. The further map to the UHP is standard.

• Could you elaborate on the meaning of ramification point? Also I'm not sure if you are allowed to take square root when your region covers the origin. – Mark Jan 6 '15 at 4:52
• It’s been a while, but I’ll try to answer your questions. A ramification point, such as zero is for the map $z\mapsto z^2$, is one where, when you go around the point once, the image point makes a double circuit. It should happen then that you can take a square root, in effect, in this example, converting $z^2$ back to $z$. In any case, try the final formula, and see whether it does what it’s been advertised to do. – Lubin Jan 6 '15 at 20:51

Multiply by $-\sqrt{i}$ which is a rotation by $-45^\circ$ and then multiply by $1/2$ to make the hyperbola satisfy $x^2 - y^2 = 1/2$

Then note that this your region is exactly the image of $\sin z$ where $z \in (-\pi/4, \pi/4)$ so take your conformal map to be: $\arcsin(-1/2 \sqrt{i} z)$

Then mapping the strip to the upper half plane can be done by scaling and vertically translating the strip and taking exponential.

http://www.webassign.net/zillengmath4/20.2.pdf has a derivation plus a nice picture (Figure 20.2.2)