Conformal Mapping?

1. Sep 12, 2012

chill_factor

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?

2. Sep 13, 2012

AlephZero

Right. An analytic function is differentiable everywhere by definition.