Suppose that $z\_n,z\in G=\mathbb{C}-\{z:z\leq 0\}$ and $z\_n=r\_ne^{i\theta \_n}, z=re^{i\theta}$ where $-\pi<\theta,\theta\_n<\pi$. Prove that if $z\_n\to z$ then $\theta\_n\to \theta$ and $r\_n\to r$.

**My approach:** 

1) Since $z\_n\to z$ then $|z\_n|\to |z|$ as $n\to \infty$. Hence $|r\_n|\to |r|$ so $r\_n\to r$ as $n\to\infty$ because $r\_n,r>0$.

2) Since $\lim z\_n=z $ $\Rightarrow$ $\lim r\_ne^{i\theta\_n}=re^{i\theta}$ $\Rightarrow$ $\lim r\_n\cdot \lim e^{i\theta\_n}=re^{i\theta}$ $\Rightarrow$ $\lim e^{i\theta\_n}=e^{i\theta}$.

How to show that it implies that $\theta\_n\to \theta$? I suppose that the function $f(z)=\arg(z)$ is continuous on $\mathbb{C}-\{z:z\leq 0\}$. But I cannot prove it at all.

Qustion 1: Can anyone show the detailed proof that this is continuous function on given domain?

Question 2: I cannot understand it but can anyone explain why do we cut complex plane along negative real axis? Why it's so important?

P.S. I have seen some duplicates if these question and on those topics they are already using the continuity of $\arg(z)$. However, my question is why it is continuous and why do we cut complex plane in the above way?

Would be very grateful for detailed answer and help!