Complex Analysis: Show that $Arg(z)$ is discontinuous at each point on the nonpositive real axis.

Question

That's the question. I'm not entirely how to prove, per se, that $Arg(z)$ is discontinuous at every point on the nonpositive real axis. By definition, $- \pi < Arg(z) \leq \pi$, so I'm not sure what there is to show?

Answer 1

Consider for instance $-1$. When $z$ approaches $-1$ from the upper half plane, $\text{Arg}(z)$ approaches $\pi$. When $z$ approaches $-1$ from the lower half plane, $\text{Arg}(z)$ approaches $-\pi$.

Answer 2

Break up into the case $z=0$ and $z<0$. At $z=0$ take the paths $z_n=\frac1n(-1+i)$ and $w_n=\frac1n(-1-i)$. Clearly $z_n,w_n\to0$ as $n\to\infty$, but $\operatorname{Arg}z_n=\frac{3\pi}4$ and $\operatorname{Arg}w_n=-\frac{3\pi}4$, so $\operatorname{Arg}$ cannot be continuous at $z=0$.

Now take $z<0$ real. Then let $z_n=e^{-i/n}z$ and $w=e^{i/n}z$. Again it is clear that $z_n,w_n\to0$, but $\operatorname{Arg}z_n=\pi-\frac1n\to\pi$ and $\operatorname{Arg}w_n=-\pi+\frac1n\to-\pi$, so $\operatorname{Arg}$ cannot be continuous at $z$.

Answer 3

Another way to prove Let $$ f(z)=\mathrm{Arg}(z)= \begin{cases} \pi + \arctan{(y/x)}&\text{when }x<0, y\ge0\\ -\pi+ \arctan{(y/x)}&\text{when }x<0, y<0 \end{cases} $$ Now $\lim_{z\to x+i0} f(z)=-\pi$ , $x<0 y>0$ (approaching $z= x + i0$ through positive $y$ axis) and also $\lim_{z->x+i0}=\pi$ , $x<0,y<0$ (approaching $z=x+i0$ through negative $y$ axis). Then we see that the two limits are not equal so the limit does not exist so therefore $\mathrm{Arg}(z)$ is not continuous.