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

Peter

Well-known member
MHB Site Helper
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 ... ...

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
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 $$\sqrt{|z|}$$, |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 $$\displaystyle \lim_{ h \to 0+ } \theta (z_0 - ih) = - \pi$$ ... ... "

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
Because $$\sqrt{|z|}$$, |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 $$\sqrt{|z|}$$, |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 $$\displaystyle \lim_{ z \to z_0 } \sqrt{|z|} = \sqrt{|z_0|}$$ but why exactly is this true?

Can you help?

Peter