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?
Is this generally true? On an argument regarding conformal maps
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complex-analysis
taylor-expansion
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0Study 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 $az + \dots$. – 2012-02-01
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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).