Conformal Mapping?

  • #1
chill_factor
901
5
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
7,025
297
Right. An analytic function is differentiable everywhere by definition.
 

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