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!
Best Answer
$\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.