> [!NOTE] Theorem (Isometry that fixes $O$ is linear) > Let $f:\mathbb{R}^{2}\to \mathbb{R}^{2}$ be a [[Isometry of The Plane|plane isometry]] such that $f(0,0)=(0,0)$ then $f$ is either a rotation about the origin or a reflection in the $x$-axis followed by a rotation about the origin. > > Equivalently, let $f:\mathbb{C}\to \mathbb{C}$ be a function such that $f(0)=0$ and for all $w,z\in C,$ $|w-z|=|f(w)-f(z)|$ then $f$ either has the form $f(z)=e^{i\theta}$ or $f(z)=e^{i\theta}\bar{z}$ for some $\theta\in\mathbb{R}.$ > ^91f6dc *Proof*. Let $f(e_{1})=(u,v)$ and $\theta$ be the angle between the line joining $(u,v)$ and $O$ and the line joining $(1,0)$ to $O,$ measured anticlockwise from the latter to the former. Let $g_{\theta}:\mathbb{R}^{2}\to \mathbb{R}^{2}$ be the function which is an anticlockwise rotation through angle $\theta$ about the origin. This clearly has an inverse. Consider $(g_{\theta}^{-1} \circ f) (1,0) = g_{\theta}^{-1} (u,v) = (1,0) $Also $(g_{\theta}^{-1} \circ f) (0,0) = g_{\theta}^{-1}(0,0)=(0,0)$ and $g_{\theta}^{-1}\circ f$ is an isometry since both $f$ and $g_{\theta}^{-1}$ are isometries. Therefore, by the above lemma, $g_{\theta}^{-1}\circ f$ is either the identity map or a reflection in the $x$-axis. If $g_{\theta}^{-1}\circ f$ is the identity map, then $f=(g_{\theta} \circ g_{\theta}^{-1})\circ f= g_{\theta}\circ (g_{\theta}^{-1}\circ f) =g_{\theta}.$ If $g_{\theta}^{-1}\circ f$ is a reflection in the $x$-axis, then $f=(g_{\theta}\circ g_{\theta}^{-1})\circ f = g_{\theta} \circ(g_{\theta}^{-1}\circ f).$ Thus $f$ is either a rotation about $O$ or a reflection in the $x$-axis followed by a rotation about $O.$