0
$\begingroup$

In the text, Functions of one complex variable, page 124 Conway stated the following:

enter image description here

This is intuitively clear. Being an interior point the first two lines follows from the definition. Problem is at the last two sentences. Can someone elaborate it more mathematically. Thanks in advance.

New contributor
Fib1123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
0
$\begingroup$

If $\vert \xi \vert > \vert \alpha\vert $ then $\xi$ is further away from $0 \in \mathbb{C}$ than $\alpha$ is. To find such a $\xi$ consider the following construction:

If $\alpha = 0$ just take $\xi = \frac{\rho}{2} \in \mathbb{R} \subset \mathbb{C}$

If $\alpha \neq 0$: Construct the line going from the origin of the complex plane to the point $\alpha$. This line clearly intersects $B(\alpha,\rho)$ and passes through the center ($\alpha$) of the open ball. This line covers the diameter of the ball completely. Just take $\xi \in B(\alpha,\rho)$ on the diameter and further from the center than $\alpha$. For example you could take

$$\xi = \frac{\alpha + \rho\frac{\alpha}{\vert \alpha \vert}}{2}$$

There are obviously many other valid choices for $\xi$'s.This explains the third sentence.

To understand the last sentence just apply what we just did. Suppose by contradiction that $\alpha \in \Omega$ is such that $\forall \xi \in \Omega : \vert \alpha \vert \geq \vert \xi \vert $ but that $\alpha \not \in \partial \Omega$. Then $\alpha$ is in the interior of $\Omega$ and there is $\rho > 0 $ and $\xi_1 \in B(\alpha, \rho) \subset \Omega$ such that $$\vert \xi_1 \vert> \vert \alpha \vert.$$ a contradiction !

$\endgroup$

Your Answer

Fib1123 is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.