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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!

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  • $\begingroup$ 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 $[1 + i(0)].$ $\endgroup$ Commented Aug 29, 2022 at 23:19

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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 $(x_n)$.

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  • $\begingroup$ To show that the sequence $x_n$ has a limit point in $[0, \pi]$ we can use the Bolzano-Weierstrass Theorem right? $\endgroup$ Commented Aug 30, 2022 at 14:37
  • $\begingroup$ Yes, that is right. @MatteoMenghini $\endgroup$ Commented Aug 30, 2022 at 15:16
  • $\begingroup$ Why every subsequence $\{x_{n_k}\}$ of $\{x_n\}$ is convergent ? $\endgroup$
    – Winston
    Commented Jul 14, 2023 at 5:13

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