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Further Questions on Complex Square Root Function ... Palka, Example 1.5, Section 1.2, Chapter III

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
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:




Palka - 1 - Example 1.5, Section III - PART 1 ... .png
Palka - 2 - Example 1.5, Section III - PART 2 ... .png



My questions are as follows:



Question 1

In the above text by Palka 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 that makes it clear that \(\displaystyle f\), too, is continuous in \(\displaystyle D\) ... ... "

How/why exactly does the fact that \(\displaystyle \theta\) is continuous on the set \(\displaystyle D\) imply that \(\displaystyle f\) is continuous in D ... ...



Question 2

In the above text by Palka we read the following:


" ... ... we observe that \(\displaystyle \lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi\) ... ... "


Can someone please explain how/why \(\displaystyle \lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi\) ... ...



Question 3

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

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



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

Peter
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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 \(\displaystyle \theta\) is continuous on the set \(\displaystyle D = \mathbb{C} \sim ( - \infty, 0]\) (Lemma II.2.4), a fact that makes it clear that \(\displaystyle f\), too, is continuous in \(\displaystyle D\) ... ... "

How/why exactly does the fact that \(\displaystyle \theta\) is continuous on the set \(\displaystyle D\) imply that \(\displaystyle f\) is continuous in D ... …
Because [tex]\sqrt{|z|}[/tex], |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 \(\displaystyle \lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi\) ... ... "


Can someone please explain how/why \(\displaystyle \lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi\) ... …
You have dropped the condition that "\(\displaystyle z_0\) is real and negative". Since z=\(\displaystyle z_0\) is real and negative, the real part of \(\displaystyle z_0- ih\) is \(\displaystyle z_0\) and the imaginary part is \(\displaystyle -ih\), both negative numbers. We are in the fourth quadrant so the limit is \(\displaystyle -\pi\) rather than \(\displaystyle \pi\).

Question 3
In the above example Palka asserts that \(\displaystyle -i \sqrt{ \mid z_0 \mid } = - \sqrt{z_0}\) ...

Can someone please demonstrate how/why this is the case …
Again that is true because, in this case, \(\displaystyle z_0\) assumed to be "real and negative".



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

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Because [tex]\sqrt{|z|}[/tex], |z| being positive, is continuous, and the composition of continuous functions is continuous.



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


Again that is true because, in this case, \(\displaystyle z_0\) assumed to be "real and negative".





Thanks for the help, HallsofIvy ...

You write ...

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

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

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

Can you help?

Peter