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Conformal Mapping?

  1. Sep 12, 2012 #1
    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. jcsd
  3. Sep 13, 2012 #2

    AlephZero

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    Right. An analytic function is differentiable everywhere by definition.
     
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