show that this function is discontinuous at a particular point $z_1= -1$ $f(z)= \lim_{z \to -1} \arg(z)$

Question

Arg(z) is principal argument.If we simply plug in the value of -1 into function.The answer is pi.The problem is that the limit of this function doesn't exist.How is this possible?Would not the limit be pi?

Answer 1

I think you mean $f(z)=\arg z$, where $\arg z \in (- \pi, \pi].$

Let $z_n:= e^{i( \pi - \frac{1}{n})}$ and $w_n:= e^{i( -\pi - \frac{1}{n})}$. Then $z_n \to -1$ and $w_n \to -1.$

Furthermore: $f(z_n)=\pi - \frac{1}{n} \to \pi$ and $f(z_n)=-\pi - \frac{1}{n} \to - \pi$

Answer 2

If you define the $\arg z$, ($z\ne 0$) as the unique $\vartheta\in (-\pi,\pi]$, for which $z=\varrho \mathrm{e}^{i\vartheta}$ ($\varrho> 0$), then $$ \lim_{\vartheta\to 0^+} \arg (\mathrm{e}^{i(\pi-\vartheta)})=\pi \qquad\text{while}\qquad \lim_{\vartheta\to 0^+} \arg (\mathrm{e}^{i(\pi+\vartheta)})= \lim_{\vartheta\to 0^+} \arg (\mathrm{e}^{i(-\pi+\vartheta)})=-\pi $$ Hence, $\arg z$ can not be defined as a continuous function in $\mathbb C\setminus\{0\}$.

Answer 3

We take the principal value of $\arg z$, so that $-\pi < \arg z \le \pi$. So if the function $f(z)=\arg z$ is to be continuous at $z=-1$, then we must have

$$ \lim_{z \to -1} \arg z = \arg (-1) = \pi. $$

So to each $\epsilon>0$, there must correspond $\delta>0$ such that

$$ | \arg z-\pi | < \epsilon \:\:\text{whenever}\:\: |z-(-1)| < \delta. $$

Any open ball centered at $-1$ must contain points $z$ with $\Im m\,z<0$. For any such $z$, $\arg z \in (-\pi,0)$, so that $\arg z \notin (\pi-\epsilon,\pi+\epsilon)$ for $\epsilon=\pi/2$, for instance.

The existence of one $\epsilon$ to which there corresponds no $\delta$ meeting the continuity criterion shows the discontinuity of $\arg$ at $z=-1$.

In fact, essentially the same argument shows that $\lim_{z \to -r} \arg z$ does not exist, for any $r \in {\mathbb R}^+$. $\blacksquare$