How to prove $\lim_{n\to\infty}\arg (z_n) = \arg (z_0)$ 
We know that
  $$\lim_{n\to\infty} z_n = z_0$$
  means $\lim_{n\to\infty} x_n = x_0$ and $\lim_{n\to\infty} y_n = y_0$, 
  But how to prove that
  $$\lim_{n\to\infty}\arg(z_n)=\arg(z_0)$$

When we only know that $\lim_{n\to\infty} z_n = z_0$.
Thanks.
 A: Consider $z_0 \neq 0$ and away from the branch cut. Take $z_n = r_ne^{i \theta_n}$ and $z_0 = r_0e^{i \theta_0},$ where $\theta_n = \arg (z_n)$ and $\theta_0 = \arg (z_0)$ are principal value arguments.
Suppose $|z_n - z_0| < \delta.$ In this case, $z_n$ is in a disk of radius $\delta$ with center $z_0$. Using some geometry we find
$$|\theta_n - \theta_0| \leqslant \arcsin \frac{\delta}{r_0}.$$
Look at a disk with center $z_0$ and radius $δ.$ The largest deviation in $|θ_n−θ_0|$ occurs at two points on the bounding circle where the ray with angle $θ_n$ emanating from the origin is tangent. We have a right triangle with hypoteneuse $r_0$ and side $δ.$ Hence, the included angle has sine equal to $δ/r_0.$
For any $\epsilon > 0$, choose  $\delta < r_0 \sin  \epsilon.$ 
Then $|z_n - z_0| < \delta$ implies $| \theta_n - \theta_0| < \epsilon,$ and since $z_n \to z_0,$
$$\lim_{n \to \infty} \arg(z_n) = \arg(z_0).$$
This fails to converge at $z_0 = 0$,  since $\theta_n$ can assume non-zero values in any neighborhood regardless of the magnitude of $|z_n - 0| = r.$
