MHB Complex Square Root Function: Qs from Bruce P. Palka's Ex. 1.5, Ch. III

[Math Amateur]

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

I am focused on Chapter III: Analytic Functions, Section 1.2 Differentiation Rules ...

I need further help with other aspects of Example 1.5, Section 1.2, Chapter III ...

Example 1.5, Section 1.2, Chapter III, reads as follows:

View attachment 9337
View attachment 9338
My questions are as follows:
Question 1

In the above text by Palka we read the following:" ... ... Recall that the function $$\theta$$ is continuous on the set $$D = \mathbb{C} \sim ( - \infty, 0]$$ (Lemma II.2.4), a fact that makes it clear that $$f$$, too, is continuous in $$D$$ ... ... "

How/why exactly does the fact that $$\theta$$ is continuous on the set $$D$$ imply that $$f$$ is continuous in D ... ...
Question 2

In the above text by Palka we read the following:" ... ... we observe that $$\lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi$$ ... ... "Can someone please explain how/why $$\lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi$$ ... ...
Question 3

In the above example Palka asserts that $$-i \sqrt{ \mid z_0 \mid } = - \sqrt{z_0}$$ ...

Can someone please demonstrate how/why this is the case ...
Help with the above questions will be much appreciated ...

Peter

 

[HallsofIvy]

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

I am focused on Chapter III: Analytic Functions, Section 1.2 Differentiation Rules ...

I need further help with other aspects of Example 1.5, Section 1.2, Chapter III ...

Example 1.5, Section 1.2, Chapter III, reads as follows:
My questions are as follows:
Question 1

In the above text by Palka we read the following:" ... ... Recall that the function $$\theta$$ is continuous on the set $$D = \mathbb{C} \sim ( - \infty, 0]$$ (Lemma II.2.4), a fact that makes it clear that $$f$$, too, is continuous in $$D$$ ... ... "

How/why exactly does the fact that $$\theta$$ is continuous on the set $$D$$ imply that $$f$$ is continuous in D ... …

Because \sqrt{|z|}, |z| being positive, is continuous, and the composition of continuous functions is continuous.

Question 2

In the above text by Palka we read the following:" ... ... we observe that $$\lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi$$ ... ... "Can someone please explain how/why $$\lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi$$ ... …

You have dropped the condition that "[math]z_0[/math] is real and negative". Since z=[math]z_0[/math] is real and negative, the real part of [math]z_0- ih[/math] is [math]z_0[/math] and the imaginary part is [math]-ih[/math], both negative numbers. We are in the fourth quadrant so the limit is [math]-\pi[/math] rather than [math]\pi[/math].

Question 3

In the above example Palka asserts that $$-i \sqrt{ \mid z_0 \mid } = - \sqrt{z_0}$$ ...

Can someone please demonstrate how/why this is the case …

Again that is true because, in this case, [math]z_0[/math] assumed to be "real and negative".

Help with the above questions will be much appreciated ...

Peter

 

[Math Amateur]

Because \sqrt{|z|}, |z| being positive, is continuous, and the composition of continuous functions is continuous.
You have dropped the condition that "[math]z_0[/math] is real and negative". Since z=[math]z_0[/math] is real and negative, the real part of [math]z_0- ih[/math] is [math]z_0[/math] and the imaginary part is [math]-ih[/math], both negative numbers. We are in the fourth quadrant so the limit is [math]-\pi[/math] rather than [math]\pi[/math].


Again that is true because, in this case, [math]z_0[/math] assumed to be "real and negative".

Thanks for the help, HallsofIvy ...

You write ...

"Because \sqrt{|z|}, |z| being positive, is continuous, and the composition of continuous functions is continuous. ... "

Can you please show how to demonstrate that \sqrt{|z|} is continuous ...

I know we require that $$\lim_{ z \to z_0 } \sqrt{|z|} = \sqrt{|z_0|}$$ but why exactly is this true?

Can you help?

Peter