Let $\operatorname{Arg}(z)$ be principal argument function defined in branch $(-\pi, \pi]$.
Prove that $\operatorname{Arg}(z)$ is discontinuous in every point in nonpositive real axis.
"Solution":
Let $z_0$ be a point on the nonpositive part of real axis.
By $z$ approaching $z_0$ "from below" the $\operatorname{Arg}(z)$ reaches $\pi$
By $z$ approaching $z_0$ "from top" the $\operatorname{Arg}(z)$ reaches $-\pi$
Therefore for two given paths $\operatorname{Arg}(z)$ has two different limits.
Therefore $\operatorname{Arg}(z)$ has no limit.
Can you please help me find more rigorous solution which specifically bases on
$\operatorname{Arg}(z) = \arctan\left(\frac{y}{x}\right) + \pi$?
I understand that I have to find two paths for which $$\lim_{x \to x_0, \ y \to y_0}\operatorname{Arg}(z)$$ does not exist. But what are they?
Best Answer
The problem with using $\arctan(y/x)$ is that it's actually continuous across the negative real axis; its discontinuities are along the imaginary axis.
To make the branch cut you want, the definition should be like this: $$ \operatorname{Arg}(z)=\begin{cases} \arctan(y/x),\quad &x>0 \\ \arctan(y/x) + \pi,\quad &x<0, \ y>0 \\ \arctan(y/x) - \pi,\quad &x<0, \ y<0 \\ \pi/2,\quad &x=0, \ y>0 \\ -\pi/2,\quad &x=0, \ y<0 \end{cases}$$
You can check that these pieces match on both halves of the imaginary axis. Along the negative real axis, $\arctan(y/x)$ is continuous but $\pi$ clashes with $-\pi$.
As for the paths, two vertical paths will do