On the Wiki page for 'Differentiable Function' (https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_in_complex_analysis)

the part where it talks about complex differentiability, it mentions how the definition for complex differentiation is the same as for single variable real valued functions and it continues on about how it's more restrictive (i.e. Cauchy Riemann are necessary conditions for complex differentiability) and then finally how complex differentiation implies real differentiation because:

$\lim\limits_{h \to 0} \frac{\lvert f(a+h)-f(a)-f'(a)h\rvert}{\lvert h \rvert} = 0$.

But I don't see just from this expression alone how this is telling us anything about the relationship between the two in $\Bbb C$ and $\Bbb R^2$. Can anyone ELI5?

Thanks in advance


The difference has to do with the fact that for a point in $\mathbb{R}$ the limit can approach from either the right or the left, and those are the only possibilities.

That said, in $\mathbb{C}$ or $\mathbb{R}^2$ the limit can approach from infinitely many directions. In $\mathbb{C}$ we can look at limits as they approach a value from strictly "real" values or strictly "imaginary" values, in particular this is how we get the Cauchy-Riemann equations.

To answer your question we get that complex differentiation implies real differentiation because $\mathbb{R} \subset \mathbb{C}$.

  • $\begingroup$ ah I completely forgot about the the fact that real numbers are part of the complex numbers (if Im(z)=0). That makes more sense, thanks $\endgroup$ – clovis Mar 1 '19 at 3:21
  • $\begingroup$ also could elaborate a bit more on the CR equations and their relationship with all of the different paths the limit could take in $\Bbb C$ and $\Bbb R^2$? $\endgroup$ – clovis Mar 1 '19 at 3:47

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