Continuity Of Argument Function.

Question

Fix $m\in \mathbb R$.

Define $f_m :\mathbb R^2 \setminus\{(0,0)\}\rightarrow(m,m+2\pi]$ $~~$as $(x,y) \mapsto$ argument of $(x,y)$ in $(m,m+2\pi]$.

i.e $$(x,y)=\left(\cos(f_m (x,y)),\sin(f_m (x,y))\right)$$ Now if $f_m$ is continuous then its restriction to unit circle is. which is compact but its image $(m,m+2\pi]$ is not .so $f_m$ is discontinuous. And it is discontinuous on ray with angle $m$

i.e on $$\{(r\cos m,r\sin m) : r\in \mathbb R^+\}$$ and continuous on rest part of the domain. I want to prove it by mathematical argument . If $m=-\pi$ then $f_m$ is just usual Arg function . and for that i can prove because Arg function has explicit formula in terms of $\arctan$ function.

So My question Is how can we prove it for arbitrary $m\in \mathbb R$ ?

Answer 1

We know $f_m(r\cos m,r\sin m)=m+ 2\pi$ since $f_m(x,y) \in (m,m+2\pi]$, and $m<\theta \leq m+2\pi$, $f_m(r\cos \theta, r\sin \theta)=\theta$. Thus $$\lim_{\theta \rightarrow m} f_m(r\cos \theta,r\sin \theta) = \lim_{\theta \rightarrow m} \theta = m \neq m+2\pi = f_m(r\cos m,r \sin m)$$ which means that $f_m$ is discontinuous at $(r\cos m , r\sin m)$. Continuity on $\theta \in (m,m+2\pi)$ is obvious since $f_m(r\cos \theta,r\sin \theta)=\theta$ on that interval, which is continuous.