The complex multiplication is a similar map, i.e. for all $w,z_1,z_2\in \mathbb{C}$ we have $$|w||z_1 - z_2|= |wz_1-wz_2|.$$ Therefore $f:\mathbb{C}\rightarrow \mathbb{C}, f(z):=\zeta\bar\zeta^{-1}\bar{z}$ for any nonzero $\zeta\in\Bbb{C}$ must also be a isometry and since the only fixed points of $f$ are $\{a\zeta:a\in\mathbb{R}\}$, $f$ must be the the map which provides the point $z'$ which is symmetrical to $z$ on the $a\zeta$-axis.
Can somebody please explain me the last implication? That given a isometry on a plane and a set of fixed points implicates a symetry-map? I have tried to understand this Statement by drawing upon elementar-geometry books, but my linear Algebra skills are not sufficient enough. That's why I would prefer a intuitive explaination over a maybe rigorous proof in this case.
Thanks in advance.
Suppose a plane isometry $f$ fixes a line through the origin $L$ pointwise. Let $p,q\in L$ be two distinct points. For every point $z$ in the plane we have $$|p-z|=|f(p)-f(z)|=|p-f(z)| \qquad\text{ and }\qquad |q-z|=|f(q)-f(z)|=|q-f(z)|.$$ Geometrically this means that if we draw two circles centered at $p$ and $q$ and both passing through $z$, they also both pass through $f(z)$, as in the picture:
This tells us that either $f(z)=z$, or $f(z)$ is the reflection of $z$ in the line $L$. As this is true for every $z$, and $f$ is an isometry, it follows (prove this!) that $f$ is either the identity or the reflection in the line $L$.