$\newcommand{\field}[1]{\mathbb{#1}} \newcommand{\nat}{\field{N}} \newcommand{\real}{\field{R}} \newcommand{\rat}{\field{Q}} \newcommand{\cpl}{\field{C}} \renewcommand{\today}{March 7, 2013.\\Last edits: March 17, 2013.}$ {\large{\bf Isometries of the Complex Plane}}\vspace{.3in}\\ \textsc{Helmut Knaust}\medskip\\ {\footnotesize Department of Mathematical Sciences\\The University of Texas at El Paso\\El Paso TX 79968-0514}\medskip\\ {\footnotesize\today} \vspace{.4in}\\ \noindent \textbf{1. Introduction.} Isometries are distance-preserving maps. We will restrict our attention to complex functions: A function $f:\cpl\to\cpl$ is called an \textit{isometry} if it satisfies \[|f(w)-f(z)|=|w-z| \mbox{ for all } w,z\in\cpl.\] It is quite easy to check that the following examples are isometries:

1. For a fixed $z_0\in\cpl$, the function given by $f(z)= z+z_0$ \textit{translates} $z$ by $z_0$.
2. For a fixed $\theta\in\real$, the function $f(z)=z e^{i\theta}$ \textit{rotates} $z$ around the origin by the angle $\theta$.
3. The function $f(z)=\overline{z}$ \textit{reflects} $z$ about the real axis.

Isometries are injective functions. Indeed assume that $f:\cpl\to\cpl$ is an isometry, and that $w\neq z$. Then $0\neq |w-z|=|f(w)-f(z)|$, so $f(w)\neq f(z)$. It is not as easy to prove directly that isometries in the complex plane are also surjective functions, but this fact will follow from our discussion below. Isometries are used to define congruence in geometry: two geometric figures $A$ and $B$ are called \emph{congruent} to each other, if there is an isometry $f:\cpl\to\cpl$ with $f(A)=B$. \medskip \noindent \textbf{2. Classification of isometries in the complex plane.} Let $f:\cpl\to\cpl$ be an arbitrary isometry. We set $z_0=f(0)$, and then define $g:\cpl\to\cpl$ by \[g(z)=f(z)-z_0.\] The function $g$ satisfies $g(0)=0$. Next consider $g(1)$. Since $g$ is an isometry, it follows that \[|g(1)|=|g(1)-g(0)|=|1-0|=1,\] so there is a $\theta\in\real$ such that $g(1)=e^{i \theta}$. We define $h:\cpl\to\cpl$ by \[h(z)=g(z)e^{-i\theta}.\] This function satisfies the two conditions $h(0)=0$ and $h(1)=1$. Using the additional fact that $h:\cpl\to\cpl$ is an isometry, we obtain the following two equations: \begin{eqnarray} |h(z)|&=&|z|\\|h(z)-1|&=&|z-1| \end{eqnarray} Squaring both sides of (2) yields \[(h(z)-1) (\overline{h(z)-1)}=(z-1)(\overline{z-1}).\] Using the distributive property we can write this as \[h(z)\overline{h(z)}-h(z)-\overline{h(z)}+1=z \overline{z}-z -\overline{z}+1;\] in other words \[|h(z)|^2-h(z)-\overline{h(z)}+1=|z|^2-z -\overline{z}+1.\] Using (1) it follows that \[h(z)+\overline{h(z)}=z +\overline{z},\] so $h(z)$ and $z$ have the same real part. Once we know that $\operatorname{Re} h(z)=\operatorname{Re} z$, we conclude from (1) also that $\operatorname{Im} h(z)=\pm\operatorname{Im} z$. Thus $h:\cpl\to\cpl$ is given by $h(z)=z$ or by $h(z)=\overline{z}$. Working our way back we see that $g:\cpl\to\cpl$ has the form $g(z)=ze^{i\theta}$ or $g(z)=\overline{z}e^{i\theta}$, and finally we obtain that $f:\cpl\to\cpl$ is of the form \[f(z)=z_0+ze^{i\theta} \mbox{ or } f(z)=z_0+\overline{z}e^{i\theta}.\] We call these isometries \textit{of Type I}, and \textit{of Type II} respectively