Continuity of Complex Functions .... ....

[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 the proof of Lemma 2.4 ... Lemma 2.4 and its proof reads as follows:
View attachment 7358
View attachment 7359
My questions are as follows:

Question 1

In the above text from Palka Ch.2 we read in equation (2.7) that \(\displaystyle \theta (z) = - \pi - \alpha (z)\) ... ... if \(\displaystyle x \lt 0\) and \(\displaystyle y \lt 0\) ...

Now \(\displaystyle x \lt 0\) and \(\displaystyle y \lt 0\) seems to give us an angle \(\displaystyle \alpha\) in the third quadrant of the complex plane ... but \(\displaystyle ( \ - \pi - \alpha (z) \ )\) seems to give an angle in the second quadrant as we "wind both \(\displaystyle \pi\) and \(\displaystyle \alpha\) in the clockwise direction from the positive \(\displaystyle x\)-axis ... " ... so how do we get an angle in the third quadrant? ... unless it has to do with \(\displaystyle y\) being negative making \(\displaystyle \alpha\) negative ... so that \(\displaystyle - \alpha\) is positive ... ?

Can someone explain how \(\displaystyle \theta (z) = - \pi - \alpha (z)\) ... ... for \(\displaystyle x \lt 0\) and \(\displaystyle y \lt 0\) ...?***EDIT***

I have been reflecting on this question ... and have done an example calculation ... calculating \(\displaystyle \text{ Arg } z\) for \(\displaystyle z = - \sqrt{3} - i\) ... see Example calculation in the notes after the post ..

Question 2

In the above text from Palka we read the following:

" ... ... so question marks concerning the continuity of \(\displaystyle \theta\) in \(\displaystyle D\) occur only at points of the imaginary axis ... ... "Can someone please explain exactly why question marks concerning the continuity of \(\displaystyle \theta\) in \(\displaystyle D\) occur only at points of the imaginary axis ...?
Help will be appreciated ...

Peter===============================================================================

NOTES

NOTE 1

The above post refers to \(\displaystyle \alpha\) as defined in (2.6) ... ... Equation (2.6) occurs in the remarks/discussion preceding Lemma 2.4 so I am providing the relevant part of this discussion ...View attachment 7360
NOTE 2

See my calculation of \(\displaystyle \text{ Arg } z \) for \(\displaystyle z = - \sqrt{3} - i \) ... ... as follows:https://www.physicsforums.com/attachments/7361Note that there is a simple (copying) error in the above ...

Should be

\(\displaystyle -u = \frac{ \pi }{6}\)

\(\displaystyle \therefore u = - \frac{ \pi }{6}\)

Apologies ...

 

[Opalg]

... so how do we get an angle in the third quadrant? ... unless it has to do with \(\displaystyle y\) being negative making \(\displaystyle \alpha\) negative ... so that \(\displaystyle - \alpha\) is positive ... ?

I have been reflecting on this question ... and have done an example calculation ... calculating \(\displaystyle \text{ Arg } z\) for \(\displaystyle z = - \sqrt{3} - i\) ... see Example calculation in the notes after the post ..

You are correct on both counts. The definition of $\text{Arcsin}$ specifies that its domain is $[-1,1]$ and its range is $[-\pi/2,\pi/2]$. If $x$ (in that domain) is negative then so is $\text{Arcsin}\: x$. So the definition of $\alpha(z)$ ensures that if $y$ is negative then so is $\alpha(z)$.

Your calculation for $z = -\sqrt3 - i$ gives the correct result $\text{Arg}\: z = -\frac{5\pi}6$.

Can someone please explain exactly why question marks concerning the continuity of \(\displaystyle \theta\) in \(\displaystyle D\) occur only at points of the imaginary axis ...?

It should be clear from the definition that $\theta$ is continuous in the interior of each quadrant, and also on the positive real axis. It is discontinuous on the negative real axis (because it jumps between $-\pi$ and $\pi$ as it crosses the negative real axis). That only leaves the imaginary axis to deal with.

 

[Math Amateur]

You are correct on both counts. The definition of $\text{Arcsin}$ specifies that its domain is $[-1,1]$ and its range is $[-\pi/2,\pi/2]$. If $x$ (in that domain) is negative then so is $\text{Arcsin}\: x$. So the definition of $\alpha(z)$ ensures that if $y$ is negative then so is $\alpha(z)$.

Your calculation for $z = -\sqrt3 - i$ gives the correct result $\text{Arg}\: z = -\frac{5\pi}6$.It should be clear from the definition that $\theta$ is continuous in the interior of each quadrant, and also on the positive real axis. It is discontinuous on the negative real axis (because it jumps between $-\pi$ and $\pi$ as it crosses the negative real axis). That only leaves the imaginary axis to deal with.

Thanks Opalg ... really appreciate your help...

Peter