What is $\lim_{h \rightarrow 0} e^{-2i \cdot \arg(h)}$?

Question

This limit arose when I wanted to show that $f(z) = \overline{z}$ is not holomorphic.

When I say that $$\lim_{h \rightarrow 0} e^{-2i \cdot \arg(h)} = e^{-2i \cdot \arg(0)} = 1$$

I have shown that $f$ is holomorphic. So I suppose I can't simply plug in $h=0$. Why?

Answer 1

Any branch of the argument function describes the angle between a complex number and the $x$-axis. Hence the argument of $0$ is undefined, and there are points arbitrarily close to the origin for which the argument is any given number (in the range chosen according to the branch cut).

Hence $\lim_{h \to 0} \arg h$ does not exist.

Answer 2

Direct plugging is not valid. Rather, consider for example that $h=re^{i(\pi/4)}$ and $h=re^{i(\pi/3)}$ as $r\rightarrow 0^{+}$, then $e^{-2i\arg h}=e^{-2i(\pi/4)}=-1$ for the first one and $e^{-2i\arg h}=e^{-2i(\pi/3)}\ne -1$ for the second one, so the limit does not exist.

One can make any angle for that.

Another way to show that $f$ is not holomorphic is to see that $\dfrac{\partial f}{\partial\overline{z}}=\dfrac{\partial\overline{z}}{\partial\overline{z}}=1\ne 0$.