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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.

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closed as off-topic by user223391, Jack, Unit, Namaste, user99914 Feb 25 '18 at 2:07

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  • $\begingroup$ Have you learned about the Cauchy-Riemann equations yet? $\endgroup$ – Ethan Bolker Feb 24 '18 at 23:57
  • $\begingroup$ "I do not fully grasp what it means to be analytic" What definition have you read in the book? $\endgroup$ – Jack Feb 25 '18 at 1:08
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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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Every analytic function is differentiable. But $f$ isn't, that is, the limit$$\lim_{z\to0}\frac{|z|}z$$does not exist (as in the reals). So, $f$ is not analytic.

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  • $\begingroup$ 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. $\endgroup$ – Mark Viola Feb 24 '18 at 23:54
  • $\begingroup$ @MarkViola The OP did not use the expression “nowhere analytic” and what I did proves that $f$ is not analytic. $\endgroup$ – José Carlos Santos Feb 24 '18 at 23:57
  • $\begingroup$ 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). $\endgroup$ – Mark Viola Feb 24 '18 at 23:59
  • $\begingroup$ @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? $\endgroup$ – José Carlos Santos Feb 25 '18 at 0:02
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    $\begingroup$ @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? $\endgroup$ – José Carlos Santos Feb 25 '18 at 17:05
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Given $z, a, t\ne0$, with $t>0$ real, we have \begin{eqnarray} \lim_{t\to0}\frac{|z+at|-|z|}{at}&=&\lim_{t\to0}\frac{|z+at|^2-|z|^2}{at(|z+at|+|z|)}=\lim_{t\to0}\frac{t(\bar{a}z+a\bar{z})+a^2t^2}{t(|z+at|+|z|)}\\&=& \lim_{t\to0}\frac{\bar{a}z+a\bar{z}+at}{|z+at|+|z|}=\frac{\bar{a}z+a\bar{z}}{2|z|} \end{eqnarray} this shows two things: first, $f$ is not holomorphic for all $z\ne0$, and second if it were holomorphic it would have been at $z=0$. Since we also have $$ \lim_{t\to0}\frac{|0+at|-|0|}{at}=\lim_{t\to0}\frac{|a|t }{at}=\frac{|a|}{a} $$ Hence $f$ is not holomorphic.

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