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I know I can find sequences $(z_n)$, $(w_n) $ $\subset \mathbb{C}$ such that $|z_n | \to 1 $, $|w_n| \to 1 $ and

$$ \Big| \frac{ w_n - z_n}{1 - \overline{w_n} z_n } \Big| \; \; \text{does NOT converge to 1 } $$

For instance, if I take $z_n = 1 + 1/n $ and $w_n = 1 - 1/n $. However, My question is: How can I find all possible limits of such a sequences (the sequences with the required property)?? thanks

  • It is not clear exactly what you are looking for. Are you looking for limits of the pair $(w_n,z_n) \in \mathbb{C}^2$ that satisfy the above or limits of the expression $\frac{ w_n - z_n}{1 - \overline{w_n} z_n }$? – copper.hat Feb 13 '15 at 17:07
  • I suspect limits of the expression $\frac{ w_n - z_n }{1 - \overline{w_n}z_n} $ –  Feb 14 '15 at 00:15

1 Answers1

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Let $\phi(w,z) = {w-z \over 1-\bar{w}z}$.

Let $L = \{ \lim_n \phi(w_n,z_n) | |w_n| \to 1, |z_n| \to 1, \phi(w_n,z_n) \text{ is Cauchy},|\phi(w_n,z_n)| \not\to 1 \} $.

It is clear that $S^1 \cap L = \emptyset$ where $S^1 = \{z | |z|=1 \}$.

Note that $\phi(e^{i \theta} w, e^{i \theta}z) = e^{i \theta} \phi(w,z)$, hence if $\phi(w_n,z_n) \to y$, then $\phi(e^{i \theta} w_n, e^{i \theta}z_n) \to e^{i \theta} y$, hence $L = e^{i \theta} L$ for all $\theta$.

This answer shows that $[0,\infty) \setminus \{1\} \subset L$, hence $L = \mathbb{C} \setminus S^1$.

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  • Why do you assume $\phi(w_n,z_n)$ is Cauchy ? –  Feb 14 '15 at 10:50
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    If it is not Cauchy, it will not have a limit. The question asked about the limits of such sequences (see my comment to the question above). – copper.hat Feb 14 '15 at 14:45
  • I understand. However, I am still curious as to why $L = \mathbb{C} \setminus S^1 $. I see $[0, \infty) \setminus { 1 } \subset L $, but why all the complex plane except the unit circle is $L$ ? –  Feb 14 '15 at 15:07
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    Well note that I showed that $L = e^{i \theta} L$ for any $\theta$, so if $r \in L$ then so is $r e^{i \theta}$ for all $\theta$. If some point $u \in S^1$ was in $L$, this would contradict $|\phi(w_n,z_n)| \not\to 1 $ in the definition of $L$. – copper.hat Feb 14 '15 at 15:32