Conformal Mapping: Is Non-Analytic Point Conformal?

[chill_factor]

A theorem 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?

 

[AlephZero]

Right. An analytic function is differentiable everywhere by definition.