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As it says in the title, I am looking for a conformal map from $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\{z \in \mathbb{C} \mid 0 < \operatorname{Im}(z) < 7\}$, but with the following restriction on the boundary components: $(-\infty, -1]$ is mapped to $\operatorname{Im}(z) = 7$ and $[1, \infty)$ is mapped to $\operatorname{Im}(z) = 0$.

So far I have been able to map $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\{w \in \mathbb{C} \mid 0 < \operatorname{Im}(w) < 7\}$, but I don't know if the boundary components are mapped to their counterparts in the desired way. I used the following sequence of maps (note, the branch of the logarithm is always the one with argument $(0, 2\pi)$):

  • $z \mapsto\frac{1}{\sqrt{2}}\sqrt{z - 1}$ maps $\mathbb{C}\setminus((-\infty, -1]\cup[1,\infty))$ to $\mathbb{H}\setminus\{bi \mid b \in [1, \infty)\}$.

  • $z \mapsto \dfrac{z-i}{z+i}$ maps $\mathbb{H}\setminus\{bi \mid b \in [1, \infty)\}$ to $\mathbb{D}\setminus[0, 1)$.

  • $z \mapsto \sqrt{z}$ maps $\mathbb{D}\setminus[0, 1)$ to the upper half disc.

  • $z \mapsto z + \frac{1}{z}$ maps the upper half disc to the lower half plane.

  • $z \mapsto \log z$ maps the lower half plane to $\{z \in \mathbb{C} \mid \pi < \operatorname{Im}(z) < 2\pi\}$.

  • $z \mapsto \frac{7}{\pi}(z - \pi i)$ maps $\{z \in \mathbb{C} \mid \pi < \operatorname{Im}(z) < 2\pi\}$ to $\{z \in \mathbb{C} \mid 0 < \operatorname{Im}(z) < 7\}$.

The after applying the first and second map, $(-\infty, -1]$ is mapped to $[0, 1)$. What happens to this boundary component under the third map? It seems like it should remain unchanged, but geometrically it seems that it should be mapped to $(-1, 1)$.

Is there an alternative approach to this problem which would make it easier to see what happens to the boundary components?

Note, if we could construct a map as desired, except it swapped the boundary components, we could define a new map by (post)composing with the map $z \mapsto 7 - z$; this new map would then have all the desired properties.

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Michael Albanese
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    Would you mind explaining why you care? :) – Alex Youcis Aug 17 '14 at 06:03
  • @AlexYoucis: It is a past comprehensive exam question. – Michael Albanese Aug 17 '14 at 06:18
  • It depends on what you mean by "where $(0,1)$ maps under the third map": the map you've written down doesn't extend continuously to the boundary (and won't in general unless your conformal maps are between domains bounded by Jordan curves. The inverse image of your third conformal map extends continuously to the boundary; it will not be injective. So what you want to do is figure out what points have inverse image in $[0,1)$, and say that's what it maps to. This is what you call 'geometrically'. –  Aug 17 '14 at 10:02

2 Answers2

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As Mike Miller said, boundary slits should be thought of as "two-sided curves", and conformal maps are apt to split the two sides: formally, they may have different boundary limits depending on the direction of approach to the same boundary point. The simplest example of $z\mapsto \sqrt{z}$ mapping $\mathbb C\setminus (-\infty,0]$ onto half-plane: two sides of the negative semi-axis go in two different places.

For your problem, I would start with $$z\mapsto \frac{1+z}{1-z}$$ which transforms the domain into $\mathbb C\setminus (-\infty,0]$. The two boundary pieces are separated by point $-1$ (it came from $\infty$, which separated them in the original domain).

Then square root onto right halfplane; the two pieces are separated by $\pm i$: one is between $\pm i$ and one is outside. Use a fractional linear transformation to send them to $0$ and $\infty$: $$z\mapsto \frac{z+i}{z-i}$$ This was a preparation for applying the logarithm. The logarithm breaks the boundary of halfplane at $0$, sending two parts onto two sides of infinite strip. Now only cosmetics remain: moving the strip where you want it to be.

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The sine function maps the strip $\{z\in \mathbb{C}\mid |\operatorname{Re}(z)|<\tfrac{\pi}{2}\}$ biholomorphically onto $\mathbb{C}\setminus \{x\in \mathbb{R}\mid |x|\geq 1\}$. So start with $z\mapsto\arcsin(z)$ and compose with a suitable linear transformation.

WimC
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