Timeline for Generalizing contour integration to quaternions and bicomplex numbers
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2021 at 17:21 | comment | added | Anixx | @M.G. so in case of tessarines you automatically get second complex unity $ij$ and in case of bicomplex numbers you automatically get split-complex unity $ij$. | |
| Mar 23, 2021 at 17:18 | comment | added | Anixx | @M.G. what I said above is true for bicomplex numbers, tessarines and all isomorphic to them algebras. If you first add the split-complex unity $j^2=1$, you get tessarines, if you first add second complex unity $j^2=-1$, you get bicomplex numbers. But they are isomorphic to each other. | |
| Mar 23, 2021 at 17:17 | comment | added | M.G. | @Anixx: if I understand you correctly, this is just a special choice of a (nice) $\mathbb{R}$-basis for the algebra. | |
| Mar 23, 2021 at 17:11 | comment | added | Anixx | @M.G. bicomplex numbers and tessarines are 4-dimensional. One dimension is real, two are complex and the fourth dimension is hyperbolic (split-complex). So, besides complex numbers, aloso split-complex numbers are included. | |
| Mar 23, 2021 at 17:07 | comment | added | M.G. | @Anixx: as a $\mathbb{C}$-algebra, the bicomplex numbers is indistinguishable from the product algebra $\mathbb{C} \times \mathbb{C}$ because they are isomorphic as $\mathbb{C}$-algebras. Maybe you mean these algebras plus some additional structure on them that makes them distinct? | |
| Mar 23, 2021 at 16:37 | comment | added | Anixx | @M.G. well, not exactly two copies of $\mathbb{C}$. One can consider them as a combination of complex numbers and hyperbolic numbers as well. So, they are not that trivial. Yes, they have two copies of complex numbers, but also a copy of hyperbolic numbers. | |
| Mar 4, 2021 at 5:51 | comment | added | M.G. | The bicomplex numbers are indeed boring in this regard, because as an algebra they are just two copies of $\mathbb{C}$. Also, IIRC, functions that are "plane"-holomorphic when restricted to a $\mathbb{C}$-subplane of $\mathbb{H}$ are called slice holomorphic. Google returns a lot of relevant results with this search term. | |
| Mar 3, 2021 at 23:02 | history | answered | godelian | CC BY-SA 4.0 |