From $D = \{z : |z − 1| < √ 2, |z + 1| < √ 2\}$ onto the open first quadrant under a conformal mapping

Find a conformal mapping of the region $D = \{z : |z − 1| < √ 2, |z + 1| < √ 2\}$ onto the open first quadrant.

We first note that the two circles intersect at right angles.

By conformality, the images will intersect at right angles.

So if we find a LFT which maps $(−i, i, √ 2−1) → (0,∞, 1),$ we should have the required mapping.

The mapping will have the form $f(z) = α \frac{z+i}{ z−i} .$

Let $β = √ 2 − 1$ and $f(β) = α \frac{β+i}{ β−i }= 1,$ so $α = \frac{β−i}{β+i}$ and the mapping is $f(z) =\frac{ β−i}{ β+i} \frac{z+i}{ z−i}.$

This exercise and answer is from here https://www.mathstat.dal.ca/~iron/math5020/hw2sol-15.pdf Exercise 6. a)

Question:

Why finding a LFT which maps $(−i, i, √ 2−1) → (0,∞, 1),$ would be enough for the required mapping ?

Is the order of the points important? Why are the points $(0,\infty,1)$ taken in that order? I recall that I did find some transformations and the order was: $\to(1,0,\infty)$

"Why finding a LFT which maps $(−i,i,\sqrt2−1)→(0,\infty,1)$, would be enough for the required mapping?"

An LFT is conformal in any case, so it is enough to verify the boundary of the image $I$.

The LFT maps a circle/line to a circle/line. So, the images of the circles $|z-1|=\sqrt2$ and $|z+1|=\sqrt2$ are circles or lines. The two corners $\pm i$ are mapped to $0$ and $\infty$, so those boundary curves pass through $\infty$, indicating that they are some straight lines passing through $0$.

The LFT preserves orientated angles; in particular, the right angle at $-i$ is mapped to a right angle at $0$; so the two boundary lines of $I$ are perpendicular. Therefore, $I$ is a right angled angle domain with vertex at $0$.

Finally, the boundary point $\sqrt2-1$ lies on the circle arc on right side of the first domain, and it is mapped to $1$, so $1$ lies on the lower leg of that angle domain. Tis fixes the direction for the right angle.

"Is the order of the points important? Why are the points $(0,\infty,1)$ taken in that order? I recall that I did find some transformations and the order was: $\to(1,0,\infty)$"

There are two questions here.

The first question is the orientation of the points. In case of such simple regions, a conformal map sends the oriented boundary of one region to the oriented boundary of the other region. So, the order of the 3 points around the boundary must be preserved. This accords to the argument principle: if you walk around $0$ along the boundary of the first region, your image should go $+1$ times around the image of $0$. So you cannot replace say $(0,\infty,1)$ by $(\infty,0,1)$.

The second question is about cyclic shifts of the points. In general, it is possible to replace $(0,\infty,1)$ by $(1,0,\infty)$ and obtain another conformal map, but the new map will not be an LFT, because the angles at the boundary are not preserved: the right angle at $-i$ is mapped to a straight angle at $1$.

• I have a question: $i$ is sent to $\infty$, though $\infty$ does not have a right angle, why? – Isabella Sep 13 '18 at 22:34
• Could you also explain this with more detail: if you walk around $0$ along the boundary of the first region, your image should go $+1$ times around the image of $0.$ ? – Isabella Sep 13 '18 at 22:39
• Take a point form the angle domain, say let $a=1+i$. The conformal map attains $a$ exactly once with multiplicity $1$. The geometric interpretation of the argument principle is that the winding number of the boundary, with respect to $a$, is $+1$. – user141614 Sep 15 '18 at 8:08
• The angle domain has a right angle at infinity. – user141614 Sep 15 '18 at 8:11

Is the order of the points important?

No. I'll explain in a second.

Why are the points (0,∞,1) taken in that order?

The order of (0,∞,1) is dependent (in this case) on the order of (−i,i,√2−1). However, that order is indeed arbitrary. The notation that the solution uses can be a bit confusing I admit. It is simply telling you that the LFT should map: $-i \rightarrow 0$, $i \rightarrow \infty$, and $\sqrt{2}-1 \rightarrow 1$. The caveat to this is that this mapping is not strict either. For instance, the following would still be a valid mapping: $i \rightarrow 0$, $-i \rightarrow \infty$, and $-\sqrt{2}+1 \rightarrow 1$ However this different mapping would require a different transformation.. do you see why?

Why finding a LFT which maps (−i,i,√2−1)→(0,∞,1), would be enough for the required mapping?

I suspect you are new to conformal mappings? Totally fine! The mappings to $0$ and $\infty$ should be relatively self-explanatory because we are trying to map to all of the first quadrant. Mapping to $1$ might be a little less intuitive, but we are really just mapping to a finite, non-zero point. Doesn't have to be $1$ but it makes the derivation simple! This allows us (and if you don't believe me plug in a couple other points and check) to map to all of the first quadrant using the transformation that the proof goes on to derive.

Why finding a LFT which maps (−i,i,√2−1)→(0,∞,1), would be enough for the required mapping ?

Because you only need to take 3 boundary points (from the domain region) and to see the image of them in the boundary under LFT. And this would be enough to determine the complete transformation.

Is the order of the points important?

Yes.

Why are the points (0,∞,1) taken in that order?