Conformal Mapping?

  • #1
901
3

Main Question or Discussion Point

A theorm I took down in class says:

Consider the analytic function f(z). The mapping w=f(z) is conformal at the point z0 if and only if df/dz at z0 is non-zero.

However, if df/dz does not exist at that point z0, is that point still a conformal mapping? That would make the function non-analytic and this wouldn't apply right?
 

Answers and Replies

  • #2
AlephZero
Science Advisor
Homework Helper
6,993
291
Right. An analytic function is differentiable everywhere by definition.
 

Related Threads for: Conformal Mapping?

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
7
Views
4K
Top