Continuity of Complex Functions ....Palka, Example 1.5, Chapter 3 .... ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 3: Analytic Functions ...

I need help with some aspects of Example 1.5, Chapter 3 ...

Example 1.5, Chapter 3 reads as follows:View attachment 7390
View attachment 7391
In the above text from Palka Chapter 3, Section 1.2 we read the following:

" ... ... Recall that the function \(\displaystyle \theta\) is continuous on the set \(\displaystyle D = \mathbb{C} \sim ( - \infty, 0 ]\) (Lemma II.2.4), a fact which makes it clear that \(\displaystyle f\), too, is continuous in \(\displaystyle D\). ... ... "Now \(\displaystyle f\), I suspect, would be continuous because \(\displaystyle e^{i \theta / 2 }\) is continuous on \(\displaystyle D\) if \(\displaystyle \theta\) is continuous on the set \(\displaystyle D \)... (is that right?) and also \(\displaystyle h(z) = \sqrt{ \lvert z \rvert }\) is continuous ...

I have tried to rigorously prove that \(\displaystyle h(z) = \sqrt{ \lvert z \rvert }\) is continuous on \(\displaystyle D\) ... but failed ... did not get further than writing out the definition of continuity of a complex function in terms of \(\displaystyle \epsilon, \delta\) ...

Can someone show me how to rigorously demonstrate that \(\displaystyle h(z) = \sqrt{ \lvert z \rvert }\) is continuous on \(\displaystyle D\) ... ?Help will be much appreciated ...

Peter

 

[jvicens]

Hi Peter,

First of all, it is great that you are reading Palka's book on complex function theory. It is a fundamental and important subject in mathematics.

To answer your question, yes, your reasoning is correct. Since \theta is continuous on D, and e^{i \theta / 2} and \sqrt{\lvert z \rvert} are both continuous functions, it follows that f(z) = e^{i \theta / 2} \cdot \sqrt{\lvert z \rvert} is also continuous on D.

To prove the continuity of h(z) = \sqrt{ \lvert z \rvert } on D, we can use the definition of continuity as you mentioned. Let z_0 be any point in D and let \epsilon > 0 be given. We want to show that there exists a \delta > 0 such that for all z \in D, if \lvert z - z_0 \rvert < \delta, then \lvert h(z) - h(z_0) \rvert < \epsilon.

Note that since D = \mathbb{C} \sim (-\infty, 0], we can choose \delta = \min \{1, \lvert z_0 \rvert \}. Then for any z \in D with \lvert z - z_0 \rvert < \delta, we have:

\begin{align*}
\lvert h(z) - h(z_0) \rvert &= \lvert \sqrt{\lvert z \rvert} - \sqrt{\lvert z_0 \rvert} \rvert \\
&\leq \lvert \sqrt{\lvert z \rvert} - \sqrt{\lvert z_0 \rvert} \rvert \cdot \frac{\lvert \sqrt{\lvert z \rvert} + \sqrt{\lvert z_0 \rvert} \rvert}{\lvert \sqrt{\lvert z \rvert} + \sqrt{\lvert z_0 \rvert} \rvert} \\
&= \frac{\lvert z - z_0 \rvert}{\lvert \sqrt{\lvert z \rvert} + \sqrt{\lvert z_0 \rvert} \rvert} \\
&\leq \frac{\