Timeline for Harmonic polynomials on complex 2-space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2012 at 12:37 | comment | added | Alexander Isaev | Michael, one example of such a polynomial is |z|4+|w|4−4|z|2|w|2+i(|z|2−|w|2). | |
| Dec 15, 2012 at 2:24 | answer | added | Robert Bryant | timeline score: 4 | |
| Dec 14, 2012 at 19:44 | comment | added | Michael Renardy | Do you have an example of such a polynomial that is a function of |z| and |w|? I note that $|z|^2-|w|^2$ does have zeros on the unit sphere. | |
| Dec 14, 2012 at 14:18 | answer | added | Alexandre Eremenko | timeline score: 1 | |
| Dec 14, 2012 at 13:58 | comment | added | Dima Pasechnik | It looks like you have harmonic polynomials on $\mathbb{R}^4$. You can compute the dimensions of the spaces of homogeneous harmonic polynomials, and I guess the condition $f(0)=0$ alone would not suffice to cut out only the functions of $|z|$ and $|w|$. | |
| Dec 14, 2012 at 11:50 | comment | added | Alexander | For example, the harmonic function $|z|^2-|w|^2$ is a function of $|z|$ and $|w|$ alone. By the way, in my question I should have probably added "after a linear change of the real variables". | |
| Dec 14, 2012 at 11:31 | comment | added | Michael Renardy | Maybe he means a polynomial in z,w,$\bar z$ and $\bar w$. | |
| Dec 14, 2012 at 11:01 | comment | added | Pietro Majer | I don't quite understand. A non-constant polynomial $f(z,w)\in\mathbb{C}[z,w]$ is never a function of $|z|$ and $|w|$ alone. Maybe you mean something different? | |
| Dec 14, 2012 at 10:08 | history | asked | Alexander | CC BY-SA 3.0 |