## A question related to the difference between non-tangential and tangential convergence [closed]

Let $D$ be the unit open disk in $C$ and $S^1$ be its boundary.

Let me start by defining $z_n\in D\to 1 \in S^1$ "non-tangentially" ( abbreviated n.t. ) as $z_n\to 1$ but all $z_n$ lie in an angular domain $A$ at $1\in S^1$, i.e. $A$ is the set of all $z\in D : | Arg ( 1 - z ) | \le \theta$ for some $\theta > 0$. You can think of $A$ as the domain bounded in $D$ by the two straight lines that make an angle $\theta$ and $-\theta$ with the radial line segment $[0,1]$.

Let $z_n = x_n + i y_n$. It is easy to show that there is$M > 0$ such that the hyperbolic distance $d_n = d_D(z_n,x_n) \le M \forall n \ge 1$. The proof is to write $e^{d_n} = \frac{1+|\frac{z_n - x_n}{1 - x_n. z_n}|}{1-|\frac{z_n - x_n}{1 - x_n. z_n} |}$. Call $a_n = |\frac{z_n - x_n}{1 - x_n. z_n}| = \frac{|y_n|}{|1- x_n.z_n|}$ .Note that $|1 - x_n. z_n|\ge |1- z_n|$ when $x_n > 0$, therefore $a_n \le \frac{|y_n|}{|1- z_n|} = sin (\theta_n) \le k < 1 \forall n$,here $\theta_n$= angle between $[1,z_n]$ and $[0,1]$. Hence $e^{d_n} \le \frac{1+k}{1-k}$ and it is proved.

But my question is : if $z_n\in D\to 1 \in S^1$ tangentially, i.e. NOT lying in ANY such angular domain $A$ as mentioned before, must we have $d_n = d_D(z_n,x_n)\to \infty$ as $n\to \infty$ ? In other words, must

$a_n = |\frac{z_n - x_n}{1 - x_n. z_n}| = \frac{|y_n|}{|1- x_n.z_n|} \to 1$ Note that to prove the non-tangential case, we used the inequality $|1 - x_n. z_n|\ge |1- z_n|$ in a suitable way for our favor, here we cannot do that .

If the limit is not infinity, can we have some tangential convergence for some sequence $z_n \to 1$ such that $d_n = d_D(z_n,x_n) \le M \forall n \ge 1$ ?? Can we have explicit example or characterization of such tangential convergences ?

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You have a simple explicit formula for everything. Do you really believe that someone else's ability to do elementary algebra with complex numbers is better than your own? Have some more confidence! Voting to close. – fedja Dec 15 2011 at 23:54
You're basically asking us a question about elementary inequalities with complex numbers, which is not research level. The necessary calculations are fairly easy but tedious, so I think this shouldn't be on MO. – Zen Harper Dec 16 2011 at 2:24
...but I don't know the answer for certain, otherwise I would give it. – Zen Harper Dec 16 2011 at 2:27
I understand that this is very far from being a research level question, sorry to ask it here, I guess the answer is yes, the distance goes to infinity for tangential convergence, but I need to do the calculation. – Analysis Now Dec 16 2011 at 4:21
If you're really stuck, you might want to ask at math.stackexchange.com – S. Carnahan Dec 16 2011 at 5:34