Problem on Complex Analysis from Conway's book 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!
 A: $\newcommand{\Re} {\operatorname{Re}}\newcommand{\Im} {\operatorname{Im}} \newcommand{\Log} {\operatorname{Log}} \newcommand{\Arg} {\operatorname{Arg}} $Method 1 (logarithms) .
You know complex logarithms? Recall the principle value logarithm which is given for all $w\in G$ by:
$$\Log(w) =\ln |w|+i\Arg(w) $$
where $\Arg(w) \in (-\pi, \pi) $. By the definition of the logarithm we know it is continuous, hence $\Im\Log(w) =\Arg(w) $ is continuous as well.
So
$$\lim_{n\to \infty} \theta_n=\lim_{n\to\infty} \Arg(z_n) =\lim_{n\to\infty} \Im\Log(z_n)=\Im\log(z) = \Arg(z) =\theta $$
We really needed to use the fact that $\theta_n, \theta\in (-\pi, \pi) $ do you see where?
For you second question, we actually don't need to cut along the negative axis, as long as we know $z\neq 0$ and that $\theta_n$ and $\theta$ are in some interval $I$ such that  $I\cap(I+2k\pi)=\emptyset$. Why? Because otherwise the $\theta_n$ and $\theta$ would not be uniquely defined.
Method 2. 
If you don't like the use of logarithms, well then try to express the argument using arctangent etc.
For instance if $z$ has the property that $\Arg(z) \in(-\pi/2,\pi/2)$ then we know  $\Arg(z_n) \in (-\pi/2,\pi/2)$ for all $n$ large enough (Why?). In that case 
$$\theta_n=\Arg(z_n) =\arctan\left(\frac{\Im(z_n) } {\Re(z_n) } \right) $$
Now we use the continuity of the arctangent to conclude. The other cases can be handled using $w_n=  e^{i\alpha} z_n$ with appropriate choice of $\alpha$...which I leave it up to you. 
A: The function $G \to \mathbb{R}_{>0}$, $z \mapsto |z|$ is continuous.
Let $\mathbb{T}^*$ denote the punctured unit circle $\{u \in G : |u| = 1, \ u \ne -1\}$.
There is a homeomorphism $G \to \mathbb{R}_{>0} \times \mathbb{T}^*$, $z \mapsto \left(|z|, \frac{z}{|z|}\right)$, with inverse $\mathbb{R}_{>0} \times \mathbb{T}^* \to G$, $(r, u) \mapsto ru$.
From the elementary properties of the sine and cosine functions of a real variable, there is a continuous function:
$$
\operatorname{cis} \colon (-\pi, \pi) \to \mathbb{T}^*, \ \theta \mapsto \cos\theta + i\sin\theta.
$$
$\mathbb{T}^*$ is the union of three open subsets (open in $\mathbb{T}^*$, that is):
\begin{align*}
A & = \{x + iy : x, y \in \mathbb{R}, \ x^2 + y^2 = 1, \ y > 0\}, \\
B & = \{x + iy : x, y \in \mathbb{R}, \ x^2 + y^2 = 1, \ x > 0\}, \\
C & = \{x + iy : x, y \in \mathbb{R}, \ x^2 + y^2 = 1, \ y < 0\}.
\end{align*}
The restrictions of $\operatorname{cis}$ to $(0, \pi)$ and $(-\pi, 0)$ are homeomorphisms onto $A$ and $C$, respectively, because $\cos$ is a homeomorphism of each of those intervals onto $(-1, 1)$, and $y = \sqrt{1 - x^2}$ in $A$, and $y = -\sqrt{1 - x^2}$ in $C$.
The restriction of $\operatorname{cis}$ to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ is a homeomorphism onto $B$, because $\sin$ is a homeomorphism of that interval onto $(-1, 1)$, and $x = \sqrt{1 - y^2}$ in $B$.
Under $\operatorname{cis}$, the partition
$\{(-\pi, 0), \{0\}, (0, \pi)\}$ of $(-\pi, \pi)$ corresponds to the
partition $\{C, \{1\}, A\}$ of $\mathbb{T}^*$. Therefore,
$\operatorname{cis}$ is bijective.  It follows that
$\operatorname{cis}$ is a homeomorphism, because each point of
$\mathbb{T}^*$ is interior to one of the open subsets $A, B, C$, on
each of which $\operatorname{cis}^{-1}$ is continuous.
Therefore, there is a homeomorphism:
$$
G \to \mathbb{R}_{>0} \times (-\pi, \pi), \ z \mapsto \left(|z|, \operatorname{cis}^{-1}\left(\frac{z}{|z|}\right)\right),
$$
whose inverse is:
$$
\mathbb{R}_{>0} \times (-\pi, \pi) \to G, \ (r, \theta) \mapsto r\operatorname{cis}\theta = re^{i\theta}.
$$
