Timeline for Visualize the phase of complex square root with complicated cut
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2023 at 14:07 | comment | added | lotus2019 | @josh What is the code for radialPlot? | |
| Aug 17, 2022 at 15:53 | history | edited | josh | CC BY-SA 4.0 | Added level curve plot |
| Aug 17, 2022 at 15:48 | comment | added | josh | @Michael E2: I added a level-curve plot to the radial branch as an edit to my code above. Maybe this is what the OP wanted. | |
| Aug 17, 2022 at 15:47 | history | edited | josh | CC BY-SA 4.0 | Added level curve plot |
| Aug 17, 2022 at 15:21 | comment | added | josh | @Michael E2: Ok, I re-read it: Looks like the OP wishes to have perhaps a ComplexContourPlot over the indicated branch as his request to replace Plot3D with ContourPlot in an earlier comment. I'll try to do that over the radial branch I created. | |
| Aug 17, 2022 at 14:51 | comment | added | Michael E2 | Yes, I see. We have different understandings of the what the OP wants. | |
| Aug 17, 2022 at 14:49 | comment | added | josh | @Michael E2: As I understand the problem, the OP wishes to extract an analytically-continuous and single-valued section or "branch" from this multi-valued funtion equal to the sector $e^{it}$ for $\theta<t<\pi-\theta$. This sector does not impinge the singular points so has no discontinuity between it and the remainder of the sheets as I show in the plot with label "Im(f) with radial branch". Perhaps thouogh this is not what the OP wishes. | |
| Aug 17, 2022 at 14:40 | comment | added | Michael E2 | And so what you drew is incorrect? Or rather, not in agreement with my understanding of the question? — I was thinking that @DuFong would clarify. I'm just surprised that there is no discontinuity along the proposed branch cut. | |
| Aug 17, 2022 at 14:21 | comment | added | josh | @Michael E2: There is no discontinuity between the radial branch $\theta<t<\pi-\theta$ as I've drawn it above and the remainder of the multivalued sheets. That radial section is simply an analytic domain of the function which does not involve any of its singular points which is equivalent to taking the disc $1+1/2 e^{it}$ out of $\sqrt{z}$ . | |
| Aug 17, 2022 at 14:09 | comment | added | Michael E2 | @DuFong, josh: I'd expect to see each plot of the real and imaginary parts to be a surface over the complex plane with a discontinuity only along the arc of the unit circle parametrized by $z=e^{it}$, $\theta \le t \le \pi-\theta$, according how I've understood the question. But I haven't yet thought of an approach that is simple enough I want to try it. — Is that the right idea, though? | |
| Aug 17, 2022 at 10:57 | history | edited | josh | CC BY-SA 4.0 | corrected typo in description of branch cuts |
| Aug 17, 2022 at 10:56 | comment | added | josh | @DuFong: Sorry, I don't understand your question. Maybe you can explain further. Also there is likely an analytic method of directly representing a desired branch by an analytic expression using perhaps $z_1=r_1e^{i t_1}$ and $z_2=r_2 e^{i t_2}$ in the expression for $f$. | |
| Aug 16, 2022 at 20:57 | comment | added | DuFong | is it possible to project to the complex plane (say, replace Plot3D by ContourPlot?) | |
| Aug 16, 2022 at 20:53 | history | edited | josh | CC BY-SA 4.0 | Added radial branch described by OP |
| Aug 16, 2022 at 17:45 | history | edited | josh | CC BY-SA 4.0 | Made more clear and easier to follow |
| Aug 16, 2022 at 16:03 | history | answered | josh | CC BY-SA 4.0 |