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I need to find sequence $(z_n) $ and $(w_n)$ such that $|z_n| \to 1 $ and $|w_n | \to 1 $ but

$$ \Big| \frac{ w_n - z_n }{1- \overline{w}_n z_n } \Big | \; \; \text{doesn't converge to} \; \; 1 $$

My try

Put $z_n = 1 + \frac{1}{n}$ and $w_n = 1 - \frac{1}{n} $, then $|z_n| = | 1 + \frac{1}{n} | \to 1 $ and $|w_n| \to 1 $, but

$$ \Big| \frac{ w_n - z_n }{1- \overline{w}_n z_n } \Big | = \Big| \frac{ - \frac{2}{n}}{1 - (1^2 - \frac{1}{n^2})} \Big| = \Big| \frac{ - \frac{2}{n} }{\frac{1}{n^2}} \Big| = 2n$$

which does not converge to $1$ as required. My question is, is this a correct solution? What are all possible limits of such sequences?

Community
  • 1

2 Answers2

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Any positive number can be a solution to your problem.

To prove this take $L\ge 0$.

If $L=1$ then, for $w_n=1-\frac{1}{n}$ and $z_n=1-\frac{1}{n^2}$, it holds $$ \Bigl|\frac{w_n-z_n}{1-\bar{w_n}z_n}\Bigr|=\frac{n^2-n}{n^2+n-1}\to 1. $$

If $L\ne 1$ then, for $w_n=1-\frac{1+L}{1-L}\frac{1}{n}$ and $z_n=1-\frac{1}{n}$, it holds $$ \Bigl|\frac{w_n-z_n}{1-\bar{w_n}z_n}\Bigr|=\frac{2L}{|2-\frac{1-L}{1+L}\frac{1}{n}|}\to L. $$

aly
  • 867
1

For a much simpler example, take

$$z_n=w_n=1$$

for all $n$. Clearly, both $|z_n|\rightarrow1$ and $|w_n|\rightarrow1$, but

$$\Big| \frac{ w_n - z_n }{1- \overline{w}_n z_n } \Big |=\Big| \frac{0}{0} \Big |$$

is undefined and in particular does not converge to $1$, as desired.