Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with cv.complex-variables
Search options not deleted
user 1004
Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
47
votes
Does a power series converging everywhere on its circle of convergence define a continuous f...
This answer is in response to final sentence, "I would be very grateful to anybody who would write a new answer with a few lines of explanation as to what Sierpinski is actually doing". In fact, it is …
- 17.1k
7
votes
Accepted
Removable sets for harmonic functions and Hardy spaces of general domains
Yes, it is true that $H^p(\Omega)$ consists only of the constant functions for all $1\le p\le\infty$. This is because all nonnegative harmonic functions $f\colon\Omega\to\mathbb{R}$ are constant.
You …
- 17.1k
15
votes
Accepted
Distribution of roots of complex polynomials
Letting $\mu_n$ be the distribution of a randomly chosen root of a random polynomial $f=c_0+c_1X+\cdots+c_nX^n$ in $\mathbb{C}[X]$ for IID random variables $c_i\in\mathbb{C}$, each chosen with some pr …
- 17.1k
11
votes
Uniformization theorem for Riemann surfaces
I learned the proof from this paper, Uniformization of Riemann Surfaces.
They actually provide three methods of proof, and I found the first of these the easiest to follow. Its not too difficult to go …
22
votes
Accepted
When are complex polynomial maps almost surjective?
Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.
As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in …
- 17.1k