By real or complex differentiable, I mean differentiable in an open set.(adsbygoogle = window.adsbygoogle || []).push({});

A simple example that distinguishes real analysis from complex analysis, is the following

Fact: f'(z) = 0 implies f is not 1-1 in any neighborhood of z.

This is not true in the real case. For [tex]f(x) = x^3[/tex], [tex]f'(0)=0[/tex], but f is 1-1 on all of R. But for the complex function [tex]f(z)=z^3[/tex], you have [tex]f'(0)=0[/tex] but f is not 1-1 in any neighborhood of 0. To see this, for any e > 0, choose x such that 0 < x < e. Then set [tex]w = x \cdot e^{i \frac{2\pi}{3}}[/tex], and you get [tex]f(w)=w^3=x^3=f(x)[/tex]. But clearly x and w are not equal, and both x,w are in B(0,e).

Anyways, I know it's a fact, now I need to prove it (the "fact"). If anyone is up late and knows the proof, then please do post it. Otherwise I will try to figure it out and post it...

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# F'(z)=0 implies f is not 1-1 near z

Loading...

Similar Threads - implies near | Date |
---|---|

I Convergence of Taylor series in a point implies analyticity | Dec 10, 2016 |

I Continuity of a function implies its existence in the neighbourhood? | Jun 16, 2016 |

Continuity at a point implies continuity in the neighborhood | Apr 18, 2015 |

Differentiability implies continuous derivative? | Jan 28, 2015 |

Asymptotic behavior of a power series near its branch point | May 1, 2013 |

**Physics Forums - The Fusion of Science and Community**