> [!NOTE] Theorem (General form of isometries) > Let $f:\mathbb{C}\to \mathbb{C}$ be an isometry then $f$ either has the form $f(z)=e^{i\theta}z+w$ or $f(z)=e^{i\theta}\bar{z}+w$for some $\theta\in \mathbb{R}$ and $w\in\mathbb{C}.$ ^e4939d *Proof*. Let $w=f(0).$ Let $h_{-w}$ be the translation through $-w,$ i.e. $h_{-w}(z)=z-w.$ Clearly $h_{-w}$ is an isometry and so $h_{-w}\circ f$ is an isometry. We have $(h_{-w}\circ f)(0)=h_{-w}(f(0))=h_{-w}(w)=w-w=0$so $h_{-w}\circ f$ is an isometry which fixes the origin. By the above theorem, either there is a real number $\theta$ such that $h_{-w}\circ f(z)=e^{i\theta}z$ for all $z\in\mathbb{C},$ which means that, for all $z\in\mathbb{C},\:f(z)-w=e^{i\theta}z \implies f(z)=e^{i\theta}z+w$, or there is a real number $\theta$ such that $h_{-w}\circ f(z)=e^{i\theta}\overline{z}$ for all $z\in\mathbb{C}$ which means that, for all $z\in\mathbb{C},\:f(z)-w=e^{i\theta}\overline{z}$, or $f(z)=e^{i\theta}\overline{z}+w.$