Continuity of the complex natural log

I want to understand the continuity of the principal branch of log(z)\log(z). Since log|z|\log|z| is already continuous, it is sufficient to show that Arg:C∖{z∈R:z≤0}→R\text{Arg} \colon \mathbb{C}\setminus \{z \in \mathbb{R}:z \le 0\} \to \mathbb{R} defined by z↦Arg(z)z \mapsto \text{Arg}(z) is continuous as well. The idea my professor mentioned was to pick a convergent sequence {zn}\{z_n\} and show that its image under Arg\text{Arg} is still convergent. Her argument goes as follows:

Fix ϵ>0\epsilon>0 and let z0z_0 be the limit point of {zn}\{z_n\}. Since the set G:=C∖{z∈R:z≤0}G:= \mathbb{C}\setminus \{z \in \mathbb{R}:z \le 0\} is open we can find a 0<δ<ϵ 0 < \delta < \epsilon such that Σδ(z0):={z∈G:|Arg(z)−Arg(z0)|<δ}⊆G. \Sigma_\delta(z_0) := \{z \in G:|\text{Arg}(z) - \text{Arg}(z_0)| < \delta \} \subseteq G. Now since zn→z0z_n \to z_0 we can find a NN so that if n≥Nn \ge N we have zn∈Σδ(z0)z_n \in \Sigma_\delta(z_0). This implies that for n≥Nn\ge N we have |Arg(zn)−Arg(z0)|<δ<ϵ|\text{Arg}(z_n) - \text{Arg}(z_0)| < \delta < \epsilon , and Arg(⋅)\text{Arg}(\cdot) preserves convergent sequences. However, it seems to me that the line "Now since zn→z0z_n \to z_0 we can find a NN so that if n≥Nn \ge N we have zn∈Σδ(z0)z_n \in \Sigma_\delta(z_0)." is precicely what we are trying to prove. What am I missing here? Additionally, what is a good way to think about convergent sequences/continuous functions in terms of polar coordinates since the representation in this way is not unique as it is in Cartesian coordinates. ================= 2   Yes, it's a petitio principii. You only know that Σδ(z0)\Sigma_{\delta}(z_0) is open when you know that the argument is continuous. In my opinion, it's much better to derive the continuity of the argument from the continuity of the logarithm. You can define the logarithm as Logz:=∫z1dζζ,\operatorname{Log} z := \int_1^z \frac{d\zeta}{\zeta}, where the integral is over any curve from 00 to zz in the slit plane. Then it's easy to see that the logarithm is continuous. – Daniel Fischer♦ 2 days ago 2   Or you can consider exp:{z:|Imz|<π}→C\exp \colon \{ z : \lvert \operatorname{Im} z\rvert < \pi\} \to \mathbb{C} - holomorphic and injective - and define the logarithm as its inverse. Once again continuity easily follows from general principles. (Inverse function theorem of real analysis for example, or the open mapping theorem.) – Daniel Fischer♦ 2 days ago ================= =================