# Why does a conformal mapping create a full tiling of semi-infinite strips in the w-plane?

I know that, specifically for linear fractional transformations, symmetric points get mapped to symmetric points.

So, if the real line gets mapped to a circle, then under a LFT, points symmetric w.r.t. the real line in the z-plane are mapped to symmetric points w.r.t. the circle in the w-plane.

But does this symmetry argument apply for other conformal mappings, like the mappings by $e^z$, $z^\alpha$, etc?

Basically, I am currently struggling a bit to understand an image plane by a conformal mapping, I think it's by $sin^{-1}(z)$. It maps the UHP to a semi-infinite strip in the w-plane. But then the book just fills out the rest of the w-plane with semi-infinite strips that are images of $sin^{-1}(z)$, and the author just says something like, "by reflection, the mapping creates a full tiling of the target plane."

Is it a symmetry argument similar (indentical?) to the one for LFT's?

So if the shaded strip in the diagram has preimage from the upper half plane, then to the left of this strip is another strip with preimage from the lower half plane, since symmetric points w.r.t. the real line are map to symmetric points with respect to image of the real line, which in this case is three edges that carve out a semi-infinite strip. Do I have the correct idea?

Here is the online reference, on page 72:

https://books.google.com/books?id=CDP1zxFJLucC&pg=PA55#v=onepage&q&f=false

Thanks,

## 1 Answer

The symmetry argument is based on the Schwarz reflection principle, which is not concerned with the structure of a map (as long as it's holomorphic), but with the geometry of the curves in which to reflect. One can reflect across line segments and circular arcs, thanks to the existence of explicit antiholomorphic involutions that fix such curves: $z\mapsto \bar z$, $z\mapsto 1/\bar z$, and their shifted/translated/rotated analogues.

In principle, one can reflect across any real-analytic arc, but this is no longer an explicit process: it essentially amounts to mapping the arc to a line segment, reflecting there, and coming back.

The exponential map and its relatives exhibit "stripey" behavior partly because they map Cartesian-aligned lines to polar-aligned lines and circles, and partly due to their periodicity. (Actually, periodicity is related to this geometry because if vertical lines are mapped to polar circles, some vertical translations become rotations by multiples of $2\pi$.)

• Hi @NormalHuman, hmm...in the Schwarz reflection principle, if f(z) is defined and analytic on the upper half of a symmetric domain, call it $\Omega^{+}$, continuous and real-valued on $R$ then define F(z) = f(z) on $\Omega^{+}$, and F(z) = $\bar{f} (\bar{z})$. Then F(z), restricted to $\Omega^{+}$ is exactly f(z), and F(z) is an analytic continuation of f(z) to the $\Omega^{-}$, which is the lower half of the symmetric domain. But, my confusion is, F(z) is not f(z); it is a piecewise-defined function that matches f(z) on the overlap (upper half of Omega). So, now to an actual example: – User001 Jul 23 '15 at 19:39
• consider my question post...then...the continuations of $sin^{-1}$ that create the tiling effect on the target plane ...are not actually the effects of $sin^{-1}$, but some variant of it, right? – User001 Jul 23 '15 at 19:40
• -- sorry for the typos in my first comment, I was thinking and editing my second comment, and ran out of time to edit the first comment :-) thanks, @NormalHuman. – User001 Jul 23 '15 at 19:47
• Holomorphic functions that agree on an open set agree everywhere. – user147263 Jul 23 '15 at 23:27