Prove that if $z_n \rightarrow z$ then $\theta_n \rightarrow \theta$ and $r_n \rightarrow r$.

Question

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 $- \pi < \theta_n,\theta< \pi$. Prove that if $z_n \rightarrow z$ then $\theta_n \rightarrow \theta$ and $r_n \rightarrow r$.

Answer 1

Since $z_n\to z$, for any $\varepsilon > 0$ there exists $N_\varepsilon$ such that for any $n\geq N_\varepsilon$, $\|z-z_n\|\leq\varepsilon$.

Now we have: $$\forall n\geq N_\varepsilon,\qquad \left|\|z\|-\|z_n\|\right|\leq \|z-z_n\|=\varepsilon, $$ hence $\|z_n\|=r_n\to r=\|z\|$.

Assuming that $\varepsilon\leq \frac{r}{2}\cdot\left|\sin\arg(z)\right|$ and that $\|z-z_n\|\leq\varepsilon$, we have: $$\left|\arg z-\arg z_n\right|\leq \arcsin\frac{\varepsilon}{r},$$ hence $\theta_n\to\theta$.