How to remember Stolz Angle correctly The Stolz angle is a condition used in Abel's Theorem:
$$|1-z|\leq M(1-|z|)$$
Q1) How do I intuitively remember (and understand this)?
Q2) In particular, is there a quick way to see that 
$$(1-|z|)\leq M|1-z|$$
is the wrong condition?
Thanks for any help.
 A: Hmm, this is a slightly "soft" question...
A Stolz angle $S$ is also known as a "non-tangential approach region", the point being that if you approach $1$ along a curve $\gamma\subset S$ then $\gamma$ is not tangent to the unit circle at $1$.
Imagine a curve $\gamma$ in the unit disk that approaches $1$, but which is tangent to the unit circle at $1$. Draw a picture. Points on that curve close to $1$ are much closer to the boundary than they are to $1$, right? Tangential approach to the boundary says $1-|z|$ is much smaller than $|1-z|$. 
So non-tangential approach says the opposite, that $1-|z|$ is not much smaller than $|1-z|$, which is to say $|1-z|\le M(1-|z|)$.
Or look at it this way: A lot of formulas are simpler for the upper half plane $y>0$. A non-tangential approach region (to the origin) in the upper half plane is defined by $$|x|<My.$$It's pretty clear I think that that defines an angle in the upper half plane, with vertex at $0$, and which is not tangent to the boundary.
Now, it's not hard to see that $|x|<My$ is equivalent to $(x^2+y^2)^{1/2}<M'y$, where $M'$ depends on $M$. And if you think about it a little, you see that (for points near the boundary point in either domain) the condition $(x^2+y^2)^{1/2}<M'y$ in the upper half plane corresponds roughly to $|1-z|<M'(1-|z|)$ in the disk.
