MHB Limits of Complex Functions .... Example from Palka ....

[Math Amateur]

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

I am focused on Chapter 2: The Rudiments of Plane Topology ...

I need help with some aspects of a worked example in Palka's remarks in Section 2.2 Limits of Functions ...

Palka's remarks in Section 2.2 which include the example read as follows:View attachment 7366
In the above text from Palka Section 2.2 we read the following:" ... ... We need only observe that for $$z \neq 0 $$

$$ \lvert ( z + 1 + z \text{ Log } z) -1 \lvert = \lvert z + z \text{ Log } z \lvert = \lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert $$

$$\le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \lvert z \lvert \lvert \text{Arg } z \lvert \le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \pi \lvert z \lvert $$ ... ...

... ... ... "
My questions relate to the above quoted equations/inequalities ... ...Question 1How does Palka get $$\lvert z + z \text{ Log } z \lvert = \lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert$$?Well ... my take ... ... Following Palka's definition of $$\text{ Log } z = \text{ ln } \lvert z \lvert + i \text{ Arg } z
$$
we get ...

$$\lvert z + z \text{ Log } z \lvert = \lvert z + z \text{ ln } \lvert z \lvert + i z \text{ Arg } z \lvert $$

BUT ...

$$\text{ln } \lvert z \lvert = \text{Log } \lvert z \lvert$$ since $$\text{Arg } \lvert z \lvert = 0$$ ... ... is that correct?

Question 2

How did Palka get

$$\lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert \le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \lvert z \lvert \lvert \text{Arg } z \lvert \le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \pi \lvert z \lvert$$ ... ...
Well ... my take ... only up to a point .. then ?

Using the extended triangle inequality $$\lvert z_1 + z_2 + z_3 \lvert \le \lvert z_1 \lvert + \lvert z_2 \lvert + \lvert z_3 \lvert$$ we get ...

$$\lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert \le \lvert z \lvert + \lvert z \text{ Log } \lvert z \lvert \lvert + \lvert i z \text{Arg } z \lvert = \lvert z \lvert + \lvert z \lvert \lvert \text{ Log } \lvert z \lvert \lvert + \lvert i z \text{Arg } z \lvert$$ ... ... (*)

But how do we proceed from here ...?

Particular worries are as follows:

(1) In (*) above after using the equation $$\lvert z w \lvert = \lvert z \lvert \lvert w \lvert $$ I get the term $$\lvert z \lvert \lvert \text{ Log } \lvert z \lvert \lvert$$ ... but Palka gets $$\lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert$$ ... how does Palka get this expression in (*) ... what explains the discrepancy between my term and Palka's ... ?

(2) How do I deal with the term $$\lvert i z \text{Arg } z \lvert$$ in order to get $$\lvert z \lvert \lvert \text{Arg } z \lvert$$ on the right hand side of the inequality as Palka does ... ? In other words how do we demonstrate that $$\lvert i z \text{Arg } z \lvert \le \lvert z \lvert \lvert \text{Arg } z \lvert$$ ... ?Help will be much appreciated ... ...Peter=======================================================================================

I believe it would be helpful for readers of the above post to have access to Palka's definition and introductory discussion of logarithms of complex numbers ... so I am providing the same ... as follows ... https://www.physicsforums.com/attachments/7367
https://www.physicsforums.com/attachments/7368

 

[Euge]

Question 1How does Palka get $$\lvert z + z \text{ Log } z \lvert = \lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert$$?Well ... my take ... ... Following Palka's definition of $$\text{ Log } z = \text{ ln } \lvert z \lvert + i \text{ Arg } z
$$
we get ...

$$\lvert z + z \text{ Log } z \lvert = \lvert z + z \text{ ln } \lvert z \lvert + i z \text{ Arg } z \lvert $$

BUT ...

$$\text{ln } \lvert z \lvert = \text{Log } \lvert z \lvert$$ since $$\text{Arg } \lvert z \lvert = 0$$ ... ... is that correct?

In the section that you posted, Palka writes "From now on we shall write $\operatorname{Log} x$ instead of $\ln x$ when $x > 0$.

Question 2

How did Palka get

$$\lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert \le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \lvert z \lvert \lvert \text{Arg } z \lvert \le \lvert z \lvert + \lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert + \pi \lvert z \lvert$$ ... ...
Well ... my take ... only up to a point .. then ?

Using the extended triangle inequality $$\lvert z_1 + z_2 + z_3 \lvert \le \lvert z_1 \lvert + \lvert z_2 \lvert + \lvert z_3 \lvert$$ we get ...

$$\lvert z + z \text{ Log } \lvert z \lvert + i z \text{ Arg } z \lvert \le \lvert z \lvert + \lvert z \text{ Log } \lvert z \lvert \lvert + \lvert i z \text{Arg } z \lvert = \lvert z \lvert + \lvert z \lvert \lvert \text{ Log } \lvert z \lvert \lvert + \lvert i z \text{Arg } z \lvert$$ ... ... (*)

But how do we proceed from here ...?

Particular worries are as follows:

(1) In (*) above after using the equation $$\lvert z w \lvert = \lvert z \lvert \lvert w \lvert $$ I get the term $$\lvert z \lvert \lvert \text{ Log } \lvert z \lvert \lvert$$ ... but Palka gets $$\lvert \lvert z \lvert \text{ Log } \lvert z \lvert \lvert$$ ... how does Palka get this expression in (*) ... what explains the discrepancy between my term and Palka's ... ?

(2) How do I deal with the term $$\lvert i z \text{Arg } z \lvert$$ in order to get $$\lvert z \lvert \lvert \text{Arg } z \lvert$$ on the right hand side of the inequality as Palka does ... ? In other words how do we demonstrate that $$\lvert i z \text{Arg } z \lvert \le \lvert z \lvert \lvert \text{Arg } z \lvert$$ ... ?

You overlooked $\lvert i \rvert = 1$ and $-\pi < \operatorname{Arg}(z) \le \pi$. Now try to proceed.

 

[Math Amateur]

In the section that you posted, Palka writes "From now on we shall write $\operatorname{Log} x$ instead of $\ln x$ when $x > 0$.

You overlooked $\lvert i \rvert = 1$ and $-\pi < \operatorname{Arg}(z) \le \pi$. Now try to proceed.

Thanks for the help, Euge ...

Working on the example now ...

Peter