Is this function continuous on the plane

Question

Let $1<p<2$ and let $\mathrm{Arg}:\mathbb R^2\backslash\{(0,0)\}\to \mathbb R^2$ be the argument function defined as (see https://en.wikipedia.org/wiki/Argument_(complex_analysis)) $$ \mathrm{Arg}(x,y)= \begin{cases} \arctan \left({\frac {y}{x}}\right),&\text{ if }x>0\\ \arctan \left({\frac {y}{x}}\right)+\pi, &\text{ if $x<0$ and $y\geq 0$} \\ \arctan \left({\frac {y}{x}}\right)-\pi, &\text{ if $x<0$ and $y< 0$} \\ \frac {\pi }{2},&\text{ if $x=0$ and $y> 0$} \\ -\frac {\pi }{2},&\text{ if $x=0$ and $y< 0$} \\ \text{undefined},&\text{ if $x=0$ and $y= 0$} \\ \end{cases} $$ Let $f(x,y)$ be defined by $$ f(x,y)=\begin{cases} (x^2+y^2)^{\frac p 2}\cos (p \mathrm{Arg(x,y)}), &\text{ if }(x,y)\neq (0,0)\\ 0, &\text{ if }(x,y)=(0,0) \end{cases}. $$ In another notation, $$ f(x,y)=\begin{cases} (x^2+y^2)^{\frac p 2}\cos (p \arctan \frac y x), &\text{ if }x>0\\ (x^2+y^2)^{\frac p 2}\cos (\frac {p \pi}2), &\text{ if }x=0\\ (x^2+y^2)^{\frac p 2}\cos (p\pi-p \arctan \frac {|y|} {|x|}), &\text{ if }x<0\\ \end{cases}. $$ My question is: is $f$ continuous and (further, harmonic) on $\mathbb R^2$?