Timeline for When is $\Re(\zeta(s)) - \Im(\zeta(s)) = 0 $ with $\Re(\zeta(s))\neq 0$ and $\Im(\zeta(s))\neq 0$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21 at 9:11 | comment | added | მამუკა ჯიბლაძე | @free_lancer At the points where $\Re(\zeta)=\Im(\zeta)=0$, the tangent line to $\Re(\zeta)=0$ and the tangent line to $\Im(\zeta)=0$ form right angle. | |
| Jan 21 at 3:54 | comment | added | MrPie | what do you mean right angle | |
| Jan 21 at 3:54 | comment | added | MrPie | yes these curves. when they meet then it create a zero or a root. It possible that one is zero but not the other. | |
| Jan 21 at 3:49 | comment | added | მამუკა ჯიბლაძე | The curves $\Re(\zeta)=0$ and $\Im(\zeta)=0$ you mean? They are sort of rotations of each other rather: where they meet (at zeros) they meet at a right angle | |
| Jan 21 at 3:42 | comment | added | MrPie | yes, they both seem to be on the line $1/2$ if I am reading them right. The real and imaginary parts are like a translation of one another. | |
| Jan 21 at 3:41 | comment | added | მამუკა ჯიბლაძე | Simultaneous zeroes of $\Re$ and $\Im$ are exactly zeroes of $\zeta$ if this is what you are asking | |
| Jan 21 at 3:40 | comment | added | MrPie | okay sorry i coudlnt see the domain wasnt quite sure. So the zeroes of real and imaginary part are both on the line $1/2$? or at least what it indicates | |
| Jan 21 at 3:38 | comment | added | მამუკა ჯიბლაძე | Trivial zeroes at $-2k$, $k\in\mathbb N$, yes. | |
| Jan 21 at 3:38 | comment | added | MrPie | this shows zeroes not on the strip? | |
| S Jan 21 at 3:35 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 4.0 | |
| S Jan 21 at 3:35 | history | made wiki | Post Made Community Wiki by მამუკა ჯიბლაძე |