MHB Complex Derivative .... Remark in Apostol, Section 16.1 .... ....

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The discussion revolves around understanding a remark in Apostol's "Mathematical Analysis" regarding a complex function that has a derivative only at the point zero in the complex plane. Participants analyze the function using the Cauchy-Riemann equations, which indicate that the equations are satisfied solely at the origin (0,0). This leads to the conclusion that the function can only have a derivative at that point. Additionally, the continuity of the functions involved supports the existence of the derivative at zero. The conversation concludes with appreciation for the clarification provided.
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I am reading Tom M Apostol's book "Mathematical Analysis" (Second Edition) ...

I am focused on Chapter 16: Cauchy's Theorem and the Residue Calculus ...

I need help in order to fully understand a remark of Apostol in Section 16.1 ...

The particular remark reads as follows:

View attachment 9279Could someone please demonstrate (in some detail) how it is the case that the complex function $$f$$ has a derivative at $$0$$ but at no other point of $$\mathbb{C}$$ ... ...Help will be much appreciated ...

Peter
 

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Hi Peter,

With $z=x+iy$, you have $f(z)=u + iv$, with $u=x^2+y^2$ and $v=0$.

What do the Cauchy-Riemann equations tell you ?
 
castor28 said:
Hi Peter,

With $z=x+iy$, you have $f(z)=u + iv$, with $u=x^2+y^2$ and $v=0$.

What do the Cauchy-Riemann equations tell you ?
Oh! Indeed ... Cauchy-Riemann equations are only satisfied at (0,0) ... therefore the only possible point where the derivative of f can exist is (0,0) ... and, given that the functions u and v ere continuous and have continuous first-order partial derivatives then f has a derivative at (0,0) ...

Thanks fir the help ... it is much appreciated ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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