MHB 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 $$\theta (z) = - \pi - \alpha (z)$$ ... ... if $$x \lt 0$$ and $$y \lt 0$$ ...

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

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

I have been reflecting on this question ... and have done an example calculation ... calculating $$\text{ Arg } z$$ for $$z = - \sqrt{3} - i$$ ... 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 $$\theta$$ in $$D$$ occur only at points of the imaginary axis ... ... "Can someone please explain exactly why question marks concerning the continuity of $$\theta$$ in $$D$$ occur only at points of the imaginary axis ...?
Help will be appreciated ...

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

NOTES

NOTE 1

The above post refers to $$\alpha$$ 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 $$\text{ Arg } z $$ for $$z = - \sqrt{3} - i $$ ... ... as follows:https://www.physicsforums.com/attachments/7361Note that there is a simple (copying) error in the above ...

Should be

$$-u = \frac{ \pi }{6}$$

$$\therefore u = - \frac{ \pi }{6}$$

Apologies ...

 

[Opalg]

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

I have been reflecting on this question ... and have done an example calculation ... calculating $$\text{ Arg } z$$ for $$z = - \sqrt{3} - i$$ ... 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 $$\theta$$ in $$D$$ 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