# Converging sequence in the unit circle

Let $$(z_n)_{n \in \Bbb Z_+}$$ be a sequence in the unit circle $$S^1$$ such that it converges to $$z_0$$. Let also $$x_n$$ and $$x_0$$ be the unique elements of $$[0,2\pi[$$ such that $$z_n = e^{ix_n}$$ and $$z_0 = e^{ix_0}$$. Now, I have a question concerning the following claims:

• $$(z_0 \neq 1) \implies (x_n \to x_0)$$
• $$(z_0 = 1) \implies$$ $$x_n$$ may be divided into two subsequences: one converging to $$0$$ and one converging to $$2\pi$$

are they true? and if so, how can one prove them?

I have tried proving them for a while now but without any success so far. My main problem is that I can not seem to prove that, for example, $$|z_n - z_0| < \epsilon$$ forces $$|x_n - x_0| < \delta$$, where $$\delta$$ depends on $$\epsilon$$ and becomes smaller as $$\epsilon$$ becomes smaller. I have also tried to represent everything geometrically by using angles, properties of circles etc., however I can not make the passage into a rigorous proof.

Any comment or answer is much appreciated and let me know if I can explain myself clearer!

• Construct the sequence of complex numbers $z_1 - z_0, z_2 - z_0, z_3 - z_0.$ In complex Analysis, if the limit of the real portions is $u$ and the limit of the imaginary portions is $v$, then the complex sequence converges to $(u + iv)$. When $z_0 \neq 1$, there will be a $\delta$-neighborhood around $z_0$ that does not cross or contain the point $[1 + i(0)].$ When $z_0$ equals $1$, then any $\delta$-neighborhood will (potentially) contain points on both sides of $[1 + i(0)].$ Commented Aug 29, 2022 at 23:19

You can prove the first part by looking at the limit points of $$(x_n)$$. Since this sequence is bounded it has limit points and it converges iff it has a unique limit point.
Suppose $$(x_{n_k})$$ converges to $$x$$. Then $$e^{ix}=e^{ix_0}$$ and we have $$0 \leq x \leq 2\pi, 0\leq x_0 <2\pi$$. Now $$x-x_0$$ must be a multiple of $$2\pi$$ and the inequalities $$0 \leq x \leq 2\pi, 0\leq x_o <2\pi$$ force $$x$$ to be equal to $$x_0$$.
Second part is similar. In this case just note that if $$a,b \in [0,2\pi]$$ and $$a-b$$ is a multiple of $$2\pi$$ implies that $$a=0, b=2\pi$$ or $$z=2\pi, b=0$$. Thus, $$0$$ and $$2\pi$$ are the only limit points of $$(x_n)$$.
• To show that the sequence $x_n$ has a limit point in $[0, \pi]$ we can use the Bolzano-Weierstrass Theorem right? Commented Aug 30, 2022 at 14:37
• Why every subsequence $\{x_{n_k}\}$ of $\{x_n\}$ is convergent ? Commented Jul 14, 2023 at 5:13