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I'm realizing how little (in some respects) I know about circles. Here's something that emerged out of something I was fiddling with. My question is whether this is

  • "well known" in the way that $229\times983=225107$ is "well known" (don't publish it unless you're publishing a table); or
  • well known in the sense that every book includes it (for suitable values of "every"); or
  • well known in the sense that everybody knows it (for at least moderately reasonable values of "everybody").

I'm looking at the circle $|z|=1$ in $\mathbb{C}$. Let $$f(z)=\dfrac{-3z+1}{z-3}.\tag{This is $f$.}$$ This of course fixes $\pm 1$ and leaves the circle invariant, and maps $\pm i$ to $\dfrac{-3\pm4i}{5}$. If we draw a circle through those two images of $\pm i$ meeting the unit circle at a right angle, it is centered at $-5/3$ and has radius $4/3$. That circle meets the real axis at $-1/3$. So look at the line $\operatorname{Re}=-1/3$. Look at the point on that line where $\operatorname{Im} = y$. Draw the line through that point and the aformentioned center $-5/3$. That line crosses the circle twice. It would seem that those two points are $f(z)$ and $f(-\bar z)$, where $z$ and $-\bar z$ are the two points on the unit circle with imaginary part $y$.

This gives us a simple geometric picture of how $f$ behaves. That allows us to use routine Euclidean geometry to show that $\theta\mapsto f(e^{i\theta})$ satisfies the differential equation $$ \left|\dfrac{dg}{d\theta}\right| = \text{constant}\cdot\operatorname{Re}\left(g-\left(-\dfrac 5 3\right)\right) $$ subject to the constraint that the values of $g$ are on the unit circle. (The equation says the rate at which $g$ moves along the circle is proportional to a certain affine function of the real part.)

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I couldn't use "linear-fractional-transformations" as a tag because it's more than 25 characters. The limit is built in to the software. –  Michael Hardy Jun 13 '12 at 23:22
    
I fixed the differential equation; it was idiotic as it stood. The left side is the rate at which $g$ moves along the circle; the right side is an affine function of the real part. –  Michael Hardy Jun 14 '12 at 2:25
    
Please do not use abbreviations in titles. –  Mariano Suárez-Alvarez Jun 14 '12 at 3:17
    
@MichaelHardy: retagging as conformal-geometry + complex-analysis since I think that already gets the point across. If you must I would also prefer mobius transformation since it is used more commonly on this site it seems. –  Willie Wong Jun 14 '12 at 8:04
    
@WillieWong : I just tried adding a "möbius-transformation" tag. I didn't go as far as trying to save the edit. But the software changed the spelling to "mobius-transformation". "Möbius" and "Moebius" are essentially the same spelling; "Mobius" is different. –  Michael Hardy Jun 14 '12 at 12:50

2 Answers 2

up vote 2 down vote accepted

In my opinion:

I don't believe your construction is "well-known" in any of your three versions of "well-known". I am pretty sure that it is the sort of thing that could appear as an exercise in any of the well-known classical texts on complex analysis, but I am pretty sure I have never done anything quite like this, despite having done many exercises from standard texts.

A very quick search through some texts (Ahlfors, Burckel, Lang, Conway) shows nothing quite like it.

I suspect (without doing any analysis yet) that your results depend on the fact that your transformation is of the form $z\mapsto\frac{z-a}{1-\bar{a}z}$, and would be keen to hear if you have investigated further.

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I haven't done much else yet. A solution of that differential equation is what I wanted. That a linear fractional transformation might do it was merely a wild guess based on the fact that there's some gross qualitative similarity between those and the thing I was looking for. –  Michael Hardy Jun 14 '12 at 1:07
    
@MichaelHardy see math.stackexchange.com/questions/123810/… –  Will Jagy Jun 14 '12 at 1:24

Your differential equation follows from the well-known properties of Fractional Linear Transformations as conformal maps.

Let me first do the specific case in your question: let $$ w = \frac{-3 z + 1}{z - 3} $$ we can solve $$ z = \frac{1+3w}{w+3} $$ and hence $$ \mathrm{d}z = \left(\frac{3}{w+3} - \frac{1+3w}{(w+3)^2} \right)\mathrm{d}w = \frac{8}{(w+3)^2} \mathrm{d}w $$ Now recall that $$ \left|\frac{\mathrm{d}w}{\mathrm{d}z}\right| = \left|\frac{(w+3)^2}{8}\right| $$ is the conformal factor of the map $f: z\mapsto w$. That is to say, a unit tangent vector based at the point $z$ on $\mathbb{R}^2$, under the conformal map $f$, becomes a tangent vector based at $w$ on $\mathbb{R}^2$ with length $\frac{\left|w+3\right|^2}{8}$.

Now, the object $\frac{\mathrm{d}g}{\mathrm{d}\theta}$ can also be written as $\partial f/\partial\theta$ and represents the image of the vector $\partial_\theta$ under the tangent map $\mathrm{d}f$. Since we are starting with $\partial_\theta$ tangent to the unit circle, it has unit norm, and hence we know that $$ \left| \frac{\mathrm{d}g}{\mathrm{d}\theta}\right| = \frac{ |g+3|^2}{8} $$ is precisely the conformal factor! Now using our knowledge that $g$ lies on the unit circle, we conclude that $$ |g+3|^2 = (\Re g + 3)^2 + (\Im g)^2 = (\Re g)^2 + 6 \Re g + 9 + 1 - (\Re g)^2 = 6 g + 10 $$ and hence we must have $$ \left|\frac{\mathrm{d}g}{\mathrm{d}\theta}\right| = \frac{3}{4}\left( g + \frac{5}{3}\right) $$ as claimed.

Note that this generalizes to any fractional linear transformation: given $$ z = \frac{a w + b}{cw + d} \implies \mathrm{d}z = \frac{ac w + ad - acw - cb}{(cw+d)^2} \mathrm{d}w= \frac{ \begin{vmatrix} a & b \\ c & d\end{vmatrix}}{(cw+d)^2}\mathrm{d}w $$

which implies that

$$ \left| \frac{\mathrm{d}w}{\mathrm{d}z} \right| = \frac{|cw + d|^2}{|ad - bc|}$$

Hence if $s\mapsto z(s)$ is a unit-speed curve such that $w(s) = f\circ z(s)$ (where $f$ is the inverse transformation to $w\mapsto z$ given above) lies on a circle centered at the origin, we have that necessarily

$$ \left| \frac{\mathrm{d}w}{\mathrm{d}s} \right| = \frac{ |c|^2|w|^2 + |d|^2 + 2 \Re(\bar{d}c w) }{|ad - bc|} $$

and in particular solves an ODE of the type "speed is an affine function of the current coordinates" (And if $\bar{d}c \in\mathbb{R}$ we can even say that the "speed is an affine function of the current $x$ coordinate").

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Based on the above, I agree with Old John that this is something that could easily appear as an exercise in a textbook or a course, albeit most likely as one in conformal or Kahler geometry, or as one in more advanced complex analysis. It is by-none-of-your-three-means well-known; but it is also something that can be intuitively explained for anyone with the correct set of backgrounds. –  Willie Wong Jun 14 '12 at 9:10
    
I'm going to print this answer today and then look at it closely. –  Michael Hardy Jun 15 '12 at 13:56
    
Your answer proves the result, but it doesn't directly answer the questions. I've given it an up-vote but accepted Old John's answer. –  Michael Hardy Jun 17 '12 at 18:08

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