What are all the angle-preserving $\mathbb{R}$-linear maps $T:\mathbb{C}\rightarrow\mathbb{C}$?

Question

I know that linear maps of the form $T(z)=\alpha z,z\in\mathbb{C}$ preserve angles between two complex numbers. However, I know that every $\mathbb{R}$-linear map can be uniquely written as $T(z)=\alpha z+\beta \bar{z},\text{with } \alpha,\beta\in\mathbb{C}$. How can I find the rest of the angle-preserving linear maps if there are more?

Answer 1

Given two nonzero complex numbers $z$, $w$ denote by $\>\circlearrowleft\!\!(z,w)$ the counterclockwise turning angle needed to flush $z$ with $w$. This angle is determined up to an additive integer multiple of $2\pi$, and satisfies $$\circlearrowleft\!\!(z,w)={\arg}{w\over z}\ .\tag{1}$$ A real linear map $T: \>z\mapsto\alpha z+\beta\bar z$ preserves angles if $$\circlearrowleft\!\!\bigl(T(z),T(w)\bigr)=\pm\circlearrowleft\!\!(z,w)\tag{2}$$ whenever $zw\ne0$. It is clear that for a given $T$ the choice implied by $\pm$ is constant. Using $(1)$ the condition $(2)$ can be translated into $$\arg T(w)-\arg T(z)=\pm(\arg w-\arg z)\ .\tag{3}$$ Chasing the $+$ sign we see that $\arg T(w)-\arg w=\arg T(z)-\arg z$ for all nonzero $z$ and $w$, hence $\arg{T(z)\over z}$ is constant on $\dot{\mathbb C}$. Since $$\arg{T(z)\over z}=\arg\left(\alpha+\beta{\bar z\over z}\right)$$ this clearly implies $\beta=0$.

Chasing the $-$ sign in $(3)$ we first note that $-\arg c=\arg\bar c$ for all $c\in\dot{\mathbb C}$. We therefore can conclude that in this case $\arg{T(z)\over\bar z}$ is constant on $\dot{\mathbb C}$. The same argument as before then leads to $\alpha=0$.

We therefore can say that the real linear maps of ${\mathbb C}$ that preserve angles are $T(z)=\alpha z$ and $T(z)=\alpha\bar z$ with $\alpha\ne0.$