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Is it generally true that a map is conformal at points where $f'(z)\neq 0$, why? (I saw this argument used in Kapoor's Complex Variables) And Kapoor alse seems to suggest that we can determine the magnification of the angle by finding the smallest $n$ where $f^{(n)}(z)\neq 0$. My suspicion is that it probably has something to do with Taylor expansion of the map?

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Study a map $f(z) = az$ near $z=0$ when $a \ne 0$. Then study the difference between that map and a map with Taylor series $az + \dots$. –  GEdgar Feb 1 '12 at 15:18

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if $f:\mathbb{C}\to\mathbb{C}$ is differentiable, then locally it looks like multiplication by a complex number, i.e. rotation and scaling, which preserves angles (if it is non-zero).

thinking about $f$ as a map from $\mathbb{R}^2$ to itself, the derivative at a point is of the form $$ \left( \begin{array}{cc} a&-b\\ b&a\\ \end{array} \right) $$ you can try it out and see that this preserves the angle between tangent vectors if it is non-zero (once again it is a rotation and scaling).

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