Stolz Angle and Complex Analysis

[l'Hôpital]

Homework Statement

Show that Abel's Limit Theorem holds as the complex number z approaches 1 if instead of taking the requirement $$|1 - z| \leq M|1-|z||$$, you restrict z to $$|arg(1- z)| \leq \alpha$$ where $$0 < \alpha < \frac{\pi}{2}$$

Homework Equations

The Attempt at a Solution

The latter requirement should yield us the former requirement.

So, I realize that this requires some geometry, but I was never actually taught geometry well, so I've been trying to do this strictly algebraically i.e. play with inequalities.

This is what I've done: Suppose z = x + iy, x > 1.

$$- \alpha < arg(1-z) = \arctan \frac{y}{x-1} < \alpha$$

So, let M be tan(alpha). We obtain the following:
$$y < M(x-1)$$

Which thus leads to

$$|1-z|^2 = (x-1)^2 + y^2 \leq (1+M^2)(x-1)^2$$
Letting K = sqrt(1+M^2)...

$$|1-z| = K|x-1| < K|\sqrt{x^2+y^2} - 1| = K| |z| - 1|$$

This works out well for x > 1, but if x < 1, I can't figure out a way of getting that last inequality.

Help. : (

 

[mighty2000]

Thank you for your question regarding Abel's Limit Theorem. I am a scientist and I would be happy to assist you in understanding this theorem.

Firstly, let us review the statement of Abel's Limit Theorem. It states that if the complex number z approaches 1, then the requirement |1-z| \leq M|1-|z|| holds, where M is a positive real number. This means that as z gets closer to 1, the distance between 1 and z must be less than or equal to a certain multiple of the distance between 1 and the absolute value of z.

Now, let us consider your proposed restriction, |arg(1-z)| \leq \alpha, where 0 < \alpha < \frac{\pi}{2}. This restriction also ensures that z approaches 1, as the argument of 1-z must be less than or equal to a certain angle from the positive real axis. However, we need to show that this restriction also leads to the requirement |1-z| \leq M|1-|z||.

To do this, we can use the fact that the argument of a complex number is related to its distance from the origin. Specifically, for a complex number z = x+iy, we have arg(z) = \arctan \frac{y}{x}. Using this, we can rewrite the given restriction as follows:

- \alpha < arg(1-z) = \arctan \frac{y}{x-1} < \alpha

Solving for y, we obtain:

y < (x-1)\tan\alpha

Now, we can use the triangle inequality to write:

|1-z| = |(x-1) + iy| \leq |x-1| + |iy| = |x-1| + |y|

Substituting the value of y we obtained earlier, we have:

|1-z| \leq |x-1| + |(x-1)\tan\alpha|

We can further simplify this by using the fact that |x-1| = |z| - 1. Thus, we have:

|1-z| \leq |z| - 1 + |z|\tan\alpha

Finally, we can rewrite this as:

|1-z| \leq (1+\tan\alpha)|z| - 1

Compar