# Distance preserving transformations of the complex plane

Show that the most general transformation fixing the origin and preserving distances is either a rotation, or a rotation followed by a reflection in the real axis, for a transformation $f: \mathbb{C} \to \mathbb{C}$.

I was thinking of first assuming that $f$ is analytic, then the Schwarz' Lemma will give us that $f(z)=az$ for $|a|=1$, showing that $f$ is indeed a rotation. However, how can one then show that for a general transformation $f$ will be a rotation followed by a reflection in the real axis?

Consider a transformation $$f$$ that preserves distances and fixes $$0$$. It follows that $$\vert f(1)\vert=1$$, that is $$f(1)=e^{it}$$ for some real $$t$$. Thus the function $$g:\mathbb{C}\to\mathbb{C}$$ defined by $$g(z)=e^{-it}f(z)$$ preserves distances and also satisfies $$g(0)=0$$, $$g(1)=1$$.

Now, let $$w=e^{i\pi/3}$$, the points $$0$$, $$1$$ and $$w$$ form an equilateral triangle, so their images $$g(0)=0$$, $$g(1)=1$$, and $$g(w)$$ also form an equilateral triangle. Thus, we have two cases, either $$g(w)=w$$ or $$g(w)=\bar{w}$$.

• Case 1. $$g(w)=w$$. Let $$z\in\mathbb{C}$$. Since $$g$$ preserves the two circles $$C(0,\vert z\vert)$$ and $$C( 1,\vert z-1\vert)$$, so it preserves their intersection which is $$\{z,\bar{z}\}$$, that is $$g(z)\in\{z,\bar{z}\}$$. Now, if $$z=\bar{z}$$ we get $$g(z)=z$$. So, assume that $$z\ne\bar{z}$$, if $$g(z)=\bar{z}$$ we get $$\vert z - w\vert=\vert \bar{z} - w\vert =\vert z-\bar{w} \vert$$, and this implies that $$z=\bar{z}$$, which is absurd. Therefore we must have $$g(z)=z$$ in this case. Consequently, $$f(z)=e^{it} z$$, and $$f$$ is the rotation of angle $$t$$ around the origin $$O$$.

• Case 2. $$g(w)=\bar{w}$$. We consider $$h:\mathbb{C}\to\mathbb{C}$$ defined by $$h(z)=\overline{g(z)}$$. Clearly, $$h$$ preserves distances, and fixes the points $$0$$, $$1$$ and $$w$$, so, according to the previous case $$h(z)=z$$ for every $$z$$. That is $$f(z)=e^{it}\overline{z}=\overline{e^{-it}z}$$ for every $$z$$. Which is the desired conclusion.

An intuitive way of seeing this is with basic plane geometry, and in particular triangulation.

Suppose I tell you the distance of z from 0 and from 1. That is, |z| =a and |z-1|=b. This gives two circles in the plane, centres on the real line. These circles intersect at either one point, or two, depending on whether z is real or not. If z is not real, then given a and b there are two possible points. Distance preservation implies continuity (a disc of radius r is mapped into a disc of radius r).

So given f(0), we know f(1) lies on the circle of radius 1 about f(0). From f(0) and f(1) we can triangulate every point of the complex plane, except for the ambiguity of reflection across the line through f(0) and f(1).

If we now ask where f(i) goes, of which there are two possibilities, and do the same triangulation thing with f(0) and f(i) as we did with f(0) and f(1), then we see that this uniquely specifies everything: given |z|=a, |z-1|=b and |z-i|=c, there is a unique complex number with this property.

So f(0) gives us a translation, f(1) then gives us a rotation about f(0), and f(i) then tells us whether we must reflect across the line through f(0) and f(1).

(This was written on a smartphone, so please forgive the poor notation.)

Here are some thoughts on this:

A simple rotation in the complex plane is of the form $$f(z) = Az$$ where A is a complex number. This is true because if $A = r_2 e^{i \theta_2}$ and $z = r_1 e^{i \theta_1}$ then $$f(z) = Az = r_2 e^{i \theta_2} r_1 e^{i \theta_1} = (r_1 r_2)e^{i (\theta_1 + \theta_2)}$$ or a simple $\theta_2$ rotation counterclockwise if you let $r_2 = 1$.

In order to map any particular $z_0$ point to 0 you need something like $z - z_0$ in the numerator of $f(z)$ so that $f(z_0) = 0$.

Also, we know that the most general transformation $f : \mathbb{C} \rightarrow \mathbb{C}$ is the mobious transformation (aka bilinear transformation), namely

$$f(z) = \frac{az - b}{cz -d}, \quad a,b,c,d \in \mathbb{C} \quad ad -bc \neq 0$$ The condition $ad -bc \neq 0$ ensures that $f$ is neither identically constant nor meaningless.

Since you need $z_0$ to be mapped at 0, then your map should look like $$f(z) = \frac{z - z_0}{cz - d}$$

Finally you need to argue that this is a rotation or reflection followed by rotation and you are done.

If you wanna prove that the bilinear transformation is locally 1-1 and conformal, simply argue that it has a non-zero derivative $$f'(z) = \frac{ad - bc}{(cz +d)^2} \neq 0$$ and to take it a step further $$\frac{a z_1 +b }{c z_1 + d} = \frac{a z_2 +b }{c z_2 + d}$$ which can be reduced to $$(ad - bc)(z_1 - z_2) = 0$$ we see that $z_1 = z_2$. Thus $f(z)$ is also globally 1-1.

Hope that helps :))