- #1
rudinreader
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By real or complex differentiable, I mean differentiable in an open set.
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...
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...
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