I'm reading Conway's complex analysis book and on page 46 the author defines conformal maps:
This makes me think why he didn't define conformal maps as usual, i.e., as a function with angle preserving property and that's all.
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1I think its just to avoid having to prove conformal maps are analytic and because only analytic functions really matter in the context of the book so nothing is lost. – kaiten Jan 02 '16 at 07:22
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On page 74 of Remmert's complex analysis text it is proved that real differentiable, orientation-preserving, angle-preserving maps are holomorphic with nonzero derivative. – James S. Cook Jan 02 '16 at 07:23
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I suppose the question you might ask is this: can you give an example of a angle preserving map which fails to be holomorphic. In some sense $f(z) = \bar{z}$ is an example, but, not if you insist on oriented angle-preservation. – James S. Cook Jan 02 '16 at 07:27
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@kaiten why does the existence of this limit avoid the author to prove the conformal maps are analytic? – user42912 Jan 02 '16 at 07:29
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@JamesS.Cook The definition of holomorphic functions in his book is given by $\lim_{x\to a}=\frac{f(z)-f(a)}{z-a}$ without modulus. – user42912 Jan 02 '16 at 07:31
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@user42912 indeed, so, what is the meaning of the given condition (the difference quotient with the modulus)? I think it may just be real differentiability. – James S. Cook Jan 02 '16 at 07:40
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@JamesS.Cook So, $f$ is real differentiable at a point $a$, now what? Thank you! – user42912 Jan 02 '16 at 07:45