1. What I understand is a definition of $\lim_{z \to z_0}f(z) = L$ is as follows. Is this correct?

    • 1.1.

    Let $f=u+iv$ be a complex function defined on an open deleted neighbourhood of some point $z_0=(x_0,y_0) \in \mathbb C$.

    • 1.2. I understand the above translates to

    Let $G \subseteq \mathbb C$. Let $z_0=(x_0,y_0) \in \mathbb C$. Let $f: G \to \mathbb C$. Let $U$ be an open neighbourhood of $z_0$ with $U \ \setminus \{z_0\} \subseteq G$. We have $f=u+iv$, for $u,v: G \to \mathbb R$.

    • 1.3. Then $\lim_{z \to z_0}f(z) = L$ is defined as the both of the ff are true

$$\lim_{(x,y) \to (x_0,y_0)}u(x,y)=\Re(L)$$

$$\lim_{(x,y) \to (x_0,y_0)}v(x,y)=\Im(L),$$

where the real limit definition of real multivariable applies.

  1. Apparently the above is an equivalent definition of the original definition of $\lim_{z \to z_0}f(z) = L$ which is given as something like...

    For all $\varepsilon > 0$, there exists $\delta > 0$ s.t. $|f(z)-L| < \varepsilon$ whenever $0 < |z-z_0| =$ $\sqrt{(x-x_0)^2+(y-y_0)^2}$ $< \delta$ BUT apparently...

...we don't restrict only to $z=(x,y) \in $ (either $G$ or $U$ or $U \ \setminus \ \{(z_0)\}$. I think $U$.) or something unlike in the case for real multivariable limits? What's going on? How does '$f(z)$' make sense if you don't have $z \in Domain(f)=G$ ? And then maybe this even applies to (1) above.

2.1. UPDATE 1: While I've found a textbook outside our syllabus that talks about $z \in G$ (by $G$ i mean domain of $f$), I've just checked out a textbook in our syllabus (Brown Churchill) that doesn't seem to have $z \in G$. What's going on please?

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  • UPDATE 2: the closest i think i've seen to $z \in G$ is that $z$ 'has an image w' but i don't think it necessarily means $z \in G$...or idk...i think 'has an image w' is like assuming rather than implying $z \in G$

  1. In particular, what do we get as the difference in the case of $v=0$, i.e. $f=u$, i.e. $f$ is real-valued (I think in this case (3) implies $\Im(L)=0$)? I mean to ask the difference between

$$\lim_{(x,y) \to (x_0,y_0)}u(x,y)=\Re(L)$$

$$\lim_{z \to z_0}f(z)=L$$

If there's such a difference then I think the 2 $\lim$'s are actually different like we ask for the 'complex limit' vs the 'real limit'. I guess both the complex and real limits are equal if they both exist, but apparently the complex limit has stricter requirements or something. So the only difference in output is whether they exist or not.

Note: This is the correct version of the incorrect question I asked previously.

  • 1
    $\begingroup$ If $z$ is not in the domain of $f$, then what is the truth value of $|f(z) - w_0| < \varepsilon$? $\endgroup$
    – Xander Henderson
    Oct 6 '21 at 3:24
  • 1
    $\begingroup$ They are implicitly assuming that $z$ is in the domain of $f$. $\endgroup$ Oct 6 '21 at 5:45
  • 2
    $\begingroup$ When it says "Let...$f$ be defined at all points $z$ in some deleted neighborhood of $z_0$", that implies (almost by definition) that for sufficiently small $\delta$, all $z$ satisfying $0<|z-z_0|<\delta$ are in the domain of $f$. Does that cover your remaining concern? $\endgroup$
    – Mark S.
    Oct 6 '21 at 12:05
  • 1
    $\begingroup$ Also if $z$ "has an image $w$ (under $f$)", that is exactly the same thing as $z$ being in the domain of $f$ $\endgroup$
    – Mark S.
    Oct 6 '21 at 12:07
  • 1
    $\begingroup$ @MarkS. thank you everyone. this is too weird. but anyhoo i'm fully confident now to ask my instructor about this. $\endgroup$ Oct 6 '21 at 14:18

There is no difference between the "real limit" and the "complex limit". In the definition (2), you must always take $\delta$ sufficiently small so that the punctured disc $0 < |z-z_0| < \delta$ lies entirely in the domain of $f$. This is possible because of the condition (1.1).

  • $\begingroup$ thanks Ted. this is so strange. actually my instructor didn't post the lecture notes yet for this part (or maybe i don't yet have access to the course webpage) but in class my instructor said something about how complex limits are more strict than real limits. (before i of course ask my instructor myself,) have you heard anything about this? i think i've seen some posts on stackexchange about the difference between complex and real single/multivariable derivatives in this matter but... $\endgroup$ Oct 6 '21 at 2:32
  • $\begingroup$ ... as for just limits, i wasn't able to find anything online (stackexchange or elsewhere). (i've yet to check the textbooks in our syllabus though. but in 1 textbook outside syllabus i've seen so far it says $z \in G$) $\endgroup$ Oct 6 '21 at 2:33
  • 3
    $\begingroup$ The complex plane $\mathbb{C}$ and the real plane $\mathbb{R}^2$ are isometric (that is, they are indistinguishable as metric spaces). Since the limit only cares about the metric structure, limits in the complex plane are identical to limits in the real plane. However, the derivatives in the complex and real planes are somewhat different---the complex structure is more "rigid" than the real structure, hence it is "harder" for a function to be complex-differentiable than real-differentiable. @JohnSmithKyon $\endgroup$
    – Xander Henderson
    Oct 6 '21 at 2:34
  • $\begingroup$ Ted and @XanderHenderson thanks! i posted an update. apparently in brown churchill maybe it doesn't require $z \in G$ (by $G$ i mean domain of $f$). or does it? $\endgroup$ Oct 6 '21 at 2:43
  • $\begingroup$ nvm. thank you everyone. this is too weird. but anyhoo i'm fully confident now to ask my instructor about this: Ted, @XanderHenderson et al $\endgroup$ Oct 6 '21 at 14:18

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