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 75
Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.
19
votes
Does the proof of Picard's theorem become simpler by increasing the number of points that ar...
Suppose you knew that if $f$ is entire and nonconstant, then $f$ attains all but finitely many values. Then I claim Picard's theorem follows by a completely elementary argument. To show this, suppos …
- 31.2k
12
votes
Is this a rational function?
Write $f_0(z)=\sum\frac{z^n}{2^n-1}$ and inductively define $f_k(z)=f_{k-1}(z)-\frac{1}{1-z/2^k}$. By induction, it is easy to see that $f_k$ is given by the power series $$f_k(z)=\sum\frac{z^n}{2^{k …
- 31.2k