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I'd like to investigate the common turn of phrase that conflates "angle-preserving map" with "conformal map". Let $f:\Bbb R^2\to\Bbb R^2$ be a continuous function. I'll define $f$ to be angle-preserving at some $p\in\Bbb R^2$ if $$\lim_{\substack{x,y\to0\\\langle x,y\rangle=\phi|x||y|}}\frac{\langle f(p+x)-f(p),f(p+y)-f(p)\rangle}{|f(p+x)-f(p)||f(p+y)-f(p)|}=\phi$$ for all $\phi\in\Bbb R$, where the limit is over sufficiently small nonzero $x,y\in\Bbb R^2$ satisfying $\langle x,y\rangle=\phi|x||y|$ and such that $f(p+x)\ne f(p)\ne f(p+y)$, so that the division is well-defined, and by convention the "empty limit" (when the conditions are impossible to satisfy) vacuously limits to anything (and in particular limits to $\phi$), i.e. a constant function is angle preserving.

This (admittedly complex) definition is about the weakest reasonable interpretation of "angle-preserving" that I can think of, and it doesn't require $f$ to be differentiable (or even continuous, although I'll assume that for convenience). In words it says that the angle $\angle f(p+x),f(p),f(p+y)$ is close to $\cos^{-1}\phi$ whenever $x,y$ are sufficiently small and have angle $\cos^{-1}\phi$ between them. For example, the function $f(x)=xg(x/|x|)$ where $g$ is a nonconstant continuous positive real function on the unit circle is a non-differentiable, non-conformal continuous map that is angle-preserving at $0$.

For comparison, a map is conformal at $p$ if it is differentiable at $p$ and there is a $\lambda$ such that $\langle df_p(x),df_p(y)\rangle=\lambda\langle x,y\rangle$ for all $x,y\in\Bbb R^2$, i.e. it preserves inner products up to a scale, which implies angle-preservation and length-preservation (up to an overall local scale factor).

Now, the example above shows that "angle-preserving at $p$" is strictly weaker than "conformal at $p$", but the statement I'd like to prove is that "locally angle-preserving at $p$" and "locally conformal at $p$" are equivalent, which boils down to:

If $f:\Bbb R^2\to\Bbb R^2$ is an angle-preserving continuous function, then $f$ is conformal.

This question covers the case when $f$ is additionally known to be differentiable, but I am interested in this looser interpretation of "angle-preserving".

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    $\begingroup$ Do you consider the function $z \mapsto \bar{z}$ to be conformal? It's angle-preserving according to your definition. $\endgroup$ – user180040 Oct 1 '14 at 3:36
  • $\begingroup$ @user180040 Yes I do, because this is the $\Bbb R^2$-oriented inner product which satisfies $\langle x,y\rangle=\langle y,x\rangle$, and $z\mapsto\bar z$ is an isometry. In order to make it disallow that, for a more $\Bbb C$-oriented approach, change $\Bbb R^2\mapsto\Bbb C$, "differentiable" to "complex differentiable", and in the angle-preserving definition, change $\frac{\langle x,y\rangle}{|x||y|}$ to $\operatorname{arg}(x/y)$. I'll wager that the conjectured theorem is true under either interpretation - it doesn't change all that much. $\endgroup$ – Mario Carneiro Oct 1 '14 at 3:50

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