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I want to see a rigorous explanation why the following general fact: $$ f \text{ continuous at z} \Longleftrightarrow \left( \forall (x_n)_{n \in \mathbb{N}} : \lim_{n \to \infty} x_n = z \implies \lim_{n \to \infty} f(x_n) = f(z) \right) ,$$ implies that we can "approach the limit from any direction" in the case of functions $f: \mathbb{C} \to \mathbb{C}$. Specifically, I want to see why the following equivalence (last two equalities) holds: $$ f'(z) := \lim_{\substack{h \to 0 \\ h \in \mathbb{C}}} \left( \frac{f(z + h) - f(z)}{h} \right) = \lim_{\substack{h \to 0 \\ h \in \mathbb{R}}} \left( \frac{f(z + h) - f(z)}{h} \right) = \lim_{\substack{h \to 0 \\ h \in \mathbb{R}}} \left( \frac{f(z + ih) - f(z)}{ih} \right) .$$

Moreover, why should the last equality above imply that the limit is the same from all the infinitely-many other directions $z$ can be approached from (i.e., why is the Cauchy-Riemann criterion sufficient for defining differentiability)?

Abced Decba
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1 Answers1

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They’re not equivalent. The correct statement is if the first limit exists, then it is equal to the other two. Merely having Cauchy-Riemann being satisfied at a point is not enough to conclude differentiability at that point. Here’s a very obvious counterexample: define $f:\Bbb{C}\to\Bbb{C}$ such that $f(z)=0$ whenever $z$ lies on one of the coordinate axes, and define $f(z)$ in any willy-wonka fashion away from the axes (actually we don’t even need to get too crazy… just define $f=1$ away from the coordinate axes). Then, computing the last two limits in question at $z=0$ gives $0$, but the first limit doesn’t even exist. Heck, $f$ isn’t even continuous at the origin.

The correct equivalence between complex differentiability and Cauchy-Riemann have been discussed several times on this site. See for example my answer here, and the various linked related answers for more context (with slightly different hypotheses to get slightly different conclusions).

peek-a-boo
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  • Thank you for answering (refuting) my second question. Revising my first question, given that the derivative $f'(z)$ exists, I still don't see why the two other limits are equal (intuitively, yes they are "approaching horizontally and vertically", but I can't deduce this rigorously from the first fact in my post). – Abced Decba Mar 29 '23 at 17:05
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    @AbcedDecba it follows directly from what you wrote. I’m not sure what more to explain. If you want, it’s atheorem of composition of limits. – peek-a-boo Mar 29 '23 at 17:46
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    in fact even more directly, if you just write out the $\epsilon$-$\delta$ definition for limits, you'll see that the $\delta$ which 'works' for the first limit 'works' for the other same. This is the most direct and rigorous, and fastest proof there is. – peek-a-boo Mar 29 '23 at 19:33