Timeline for A question about a closed continuous curve in the complex plane
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2017 at 6:41 | vote | accept | user173856 | ||
| Sep 11, 2017 at 6:41 | comment | added | user173856 | Nate Eldredge: Thanks for your help! | |
| Sep 7, 2017 at 19:02 | answer | added | Igor Rivin | timeline score: 1 | |
| Sep 7, 2017 at 18:43 | comment | added | Nate Eldredge | @IgorRivin: Because I am too lazy to check the details :-) | |
| Sep 7, 2017 at 18:35 | comment | added | Igor Rivin | @NateEldredge Why isn't your comment an answer? | |
| Sep 7, 2017 at 17:35 | review | Close votes | |||
| Sep 8, 2017 at 8:02 | |||||
| Sep 7, 2017 at 17:31 | comment | added | user173856 | Alexandre Eremenko: I have revised my question. | |
| Sep 7, 2017 at 17:28 | history | edited | user173856 | CC BY-SA 3.0 |
added 33 characters in body
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| Sep 7, 2017 at 17:18 | comment | added | Alexandre Eremenko | Such a curve does not exist. Condition (1) implies that $\arg\gamma(t)$ is continuous on $[a,b]$ and we know that the image of a compact set under a continuous function is compact. And you have $[0,2\pi)$. | |
| Sep 7, 2017 at 16:48 | comment | added | Nate Eldredge | Intuitively, shouldn't this follow from the fact that $\gamma^n$ has winding number divisible by $n$, but a curve without self-intersections has winding number $\pm 1$? | |
| Sep 7, 2017 at 16:43 | comment | added | Nate Eldredge | Are you thinking $n \ge 2$ here? It seems to be false for $[a,b] = [0, 2\pi]$, $\gamma(t) = e^{it}$, $n=1$. (I assume you also want to rule out $t_1 = a, t_2 = b$.) | |
| Sep 7, 2017 at 16:24 | history | asked | user173856 | CC BY-SA 3.0 |