# Prove that the function $f(z)=(Arg z)^2$ is continuous in the punctured plane $\Bbb C \setminus\{0\}$

I started this by considering the cases about $Arg (z)$

$$Arg(z) = \begin{cases} \arctan(y/x), & \text{if x>0 } \\ \pi+\arctan(y/x), & \text{if x<0,y\ge 0}\\ -\pi+\arctan(y/x), & \text{if x<0,y< 0}\\ \pi/2, & \text{if x=0,y>0}\\ -\pi/2, & \text{if x=0,y<0}\\ \undefind, & \text{ifx=0,y=0} \end{cases}$$ Now I am not getting the way to go further. So please Help me to do further things.

• Do you understand that $\arg(z)$ is continuous in the plane less the negative real axis? – Mark Viola Sep 23 '15 at 4:58
• @Dr.MV I just know that $arg(z)$ is discontinuous in negative Real axis. – Chiranjeev_Kumar Sep 23 '15 at 6:52
• Yes. Say we look on the principal branch, then for $y=0^+$ and $x<0$, the argument of $z$ is $\pi$, while for $y=0^-$, it is $-\pi$. The argument function is continous elsewhere. Now, upon squaring what happens? – Mark Viola Sep 23 '15 at 13:36
• Dr. MV this become $\pi^2$ in both the cases, so it will become continuous in negative Real axis, now to finish the problem we have to show $\arg(z)$ is continuous elsewhere, can you tell me some initial lines to show this? – Chiranjeev_Kumar Sep 23 '15 at 14:53
• @Chiranjeev,have a look at my answer. – Koro Sep 23 '15 at 23:49

Let $f(z)=Arg(z)Arg(z)$ [$z\in C-\{0\}$]
$Arg(z)$ is continuous in the set: $C-\{z:Re(z)<0,Im(z)=0\}$....(A)
So, $f$ is continuous in (A).
Only problem is in the points of the set $\{z:Re(z)<0,Im(z)=0\}$....(B)
At negative $x- axis$,$\tan^{-1}(y/x)=0$;
$\lim_{y\to0{+}}Arg(z)=\pi \implies \lim_{y\to0^{+}}f(x+iy)=(Arg(z))^2=\pi^2$
$\lim_{y\to0{-}}Arg(z)=-\pi \implies \lim_{y\to0^{-}}f(x+iy)=(Arg(z))^2=(-\pi)^2=\pi^2$
$Arg(x+i0)=\pi \implies f(x)=\pi^2\implies f(x)=\lim_{y\to 0}f(x+iy)$