Prove that $f(z) = |z|$ is not analytic.

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If a function is analytic in a neighborhood of a point, then it satisfies the Cauchy-Riemann equations.

But we have

$$|z|=\sqrt{x^2+y^2}$$

Clearly, with $u=\sqrt{x^2+y^2}$ and $v=0$ we see that for $(x,y)\ne (0,0)$

$$\frac{\partial u}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}\ne 0 =\frac{\partial v}{\partial y}$$

And for $z=0$, we see that $\lim_{\Delta \to 0}\frac{|0+\Delta z|-|0|}{\Delta z}$ fails to exist.

Therefore, $|z|$ is nowhere analytic.

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Mjoseph
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Mjoseph

Updated on February 25, 2020

Comments

  • Mjoseph
    Mjoseph over 3 years

    I do not fully grasp what it means to be analytic and therefore do not know the conditions on which I have to show in order to complete the proof.

    • Ethan Bolker
      Ethan Bolker over 5 years
      Have you learned about the Cauchy-Riemann equations yet?
    • Admin
      Admin over 5 years
      "I do not fully grasp what it means to be analytic" What definition have you read in the book?
  • Mark Viola
    Mark Viola over 5 years
    You've shown only that $|z|$ is not analytic at $z=0$. You'll need to augment this to show that it is nowhere analytic.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola The OP did not use the expression “nowhere analytic” and what I did proves that $f$ is not analytic.
  • Mark Viola
    Mark Viola over 5 years
    No, you proved that $|z|$ is not analytic for $z=0$ only. It could be analytic elsewhere (it's not, but you have not shown that).
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola I don't know which definition of “analytic function” you're using, but my definition is: $f$ is analytic if, for every $z_0\in D_f$, there is a power series $\sum_{n=0}^\infty a_n(z-z_0)^n$ whose sum, in some disk $D(z_0,r)$, is equal to $f(z)$. When $D_f$ is open, this implies that $f$ is differentiable at $z_0$. What's your definition?
  • Mjoseph
    Mjoseph over 5 years
    That's the same definition given in my class. I've never heard "nowhere analytic" before and am wondering what you mean by that. As described in my class, if a function is analytic then it is analytic at every point.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @mjoseph So, do you think that I've answered your question?
  • Mjoseph
    Mjoseph over 5 years
    I'm still confused on how proving this limit doesn't exist shows that the power series doesn't equal f(z) at that point though?
  • José Carlos Santos
    José Carlos Santos over 5 years
    @mjoseph The non-existence of that limit shows that $f$ is not differentiable. And a standard fact concerning analytic functions is that they are differentiable.
  • Mjoseph
    Mjoseph over 5 years
    I was more looking for a clear cut outline of any type of proof of this type rather than just this one. I thought I would have to do something with power series since that's how it was defined and I wasn't given any examples of these proofs in class
  • Mjoseph
    Mjoseph over 5 years
    ohh okay that makes a lot of sense. So if you want to show that something isn't analytic you just have to show that it isn't differentiable at any point?
  • Mjoseph
    Mjoseph over 5 years
    or at least that is one way of showing a function isn't analytic
  • José Carlos Santos
    José Carlos Santos over 5 years
    @mjoseph It's a way of doing it, yes. And quite useful.
  • Mjoseph
    Mjoseph over 5 years
    Thank you so much for the clarification!
  • Mark Viola
    Mark Viola over 5 years
    @mjoseph The answer posted here is simply incomplete. All the user who answered has shown is that $|z|$ fails to be analytic in a neighborhood of $z=0$. Analyticity is a local (in a neighborhood) property. The function $\frac1z$ fails to be analytic in a neighborhood of $z=0$, but it is analytic on $\mathbb{C}\setminus \{0\}$. The function $|z|$ is not analytic in a neighborhood of any point. This answer fails to show that.
  • F. Conrad
    F. Conrad over 5 years
    The question is rather vague, has rather low effort and is missing context. You can not expect a fully detailed answer assuming all those missing things. I think the answer given here is perfectly fine.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola The function $\frac 1z$ neither is nor isn't analytic at $0$, since $0$ is not in its domain. And it is precisely because being analytic is a local property that, in order to prove that a function is not analytic, the only thing one has to do is to prove that it is not analytic at some point of its domain.
  • Mark Viola
    Mark Viola over 5 years
    @josecarlossantos You are conflating the term analytic (holomorphic) with the term entire. Perhaps there is a language barrier here. Do you believe that the function $1/z$ is analytic on the annulus $1<|z|<2?
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola No, I am not. An entire function is an analytice function whose domain is $\mathbb C$. At no point I assumed that.
  • Mark Viola
    Mark Viola over 5 years
    Is $1/z$ analytic on $1<|z|<2$?
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola Yes, it is.
  • Mark Viola
    Mark Viola over 5 years
    I agree. So a function can be analytic on one open domain but not analytic in another. So, is $|z|$ analytic on $1<|z|<2$? If not, how would one show that?
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola I would do as follows: if $|z|$ was analytic, its square would be too, and then I would use the Cauchy-RIemann equations to prove that this function is not differentiable and therefore not analytic.
  • Mark Viola
    Mark Viola over 5 years
    Exactly. But you never showed that $|z|$ failed to be analytic on any domain other than one that includes $0$.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola And I didn't have to. Since it cannot be represented by a power series around $0$, it is not analytic. End of proof. It's like the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=e^{-1/x^2}$ if $x\neq0$ and such that $f(0)=0$. It is not analytic because it cannot be represented by a power series near $0$. The fact that it can be represented by a power series near every other point of its domain changes nothing.
  • Mark Viola
    Mark Viola over 5 years
    @JoséCarlosSantos Again, you never showed that the function $|z|$ was not analytic on $1<|z|<2$ for example. You showed that it isn't analytic in a neighborhood of $0$. But I grow weary of this conversation.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola I think I know what's the source of our difficulty in understanding each other. When I see “Let $f(z)=$” followed by an analytical expression, I always assume, if nothing else is said about the domain, that the domain of $f$ is the set of all points at which that expression is defined. So, I assumed that the domain of $f(z)=|z|$ was $\mathbb C$. It seems to me that, as far as you are concerned, $f(z)=|z|$ is not one function. It's rather the set of all functions from a domain into $\mathbb C$ whose analytical expression is $z\mapsto|z|$. Am I right?
  • Mark Viola
    Mark Viola over 5 years
    If the domain isn't stated, then I don't assume that it is all of $\mathbb{C}$. If we were in the reals, and one asked whether $|x|$ is differentiable, I wouldn't answer "no," simply because it is not differentiable at $0$.
  • José Carlos Santos
    José Carlos Santos over 5 years
    @MarkViola I'me glad that you understand each other at last. Of course, I would say that $|x|$ is a non-differentiable function.