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.
Best Answer
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.