The Joukowski map f(z)=z+z−1 is holomorphic on C∖{0}. A holomorphic function is locally conformal where its derivative is nonzero. Here f′(z)=1−z−2 vanishes at z=±1. Thereforewhere the exclusion of zero is because the function is undefined there. Holomorphy alone does not supply conformality at the two critical points.
holomorphic on C∖{0};conformal at every point except 0,1,−1,
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