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I understand that $f\colon z \mapsto z^{5/2}$ should not be holomorphic at the origin. After all, its third derivative diverges at the origin making it impossible to have a Taylor series. However, I cannot find a sequence $(z_n)$ that approaches zero without $$ \frac{z_n^{5/2} - 0^{5/2}}{z_n - 0} = z_n^{3/2} $$ also approaching zero.

In short, how do we reconcile the definition of holomorphism (existence of the derivative) with the existence of the Taylor series for the function $f\colon z \mapsto z^{5/2}$ at the origin?

Update

Mmm, one of the comments might have meant the following. For a point other than the origin, say $y$, any sequence $(y_n)$ approaching $y$ will give different values for $$ \frac{y_n^{5/2} - y^{5/2}}{y_n - y} $$ according to which Riemann sheet that is evaluated. Since $y$ can be infinitesimally close to the origin, then $f$ is not differentiable at the origin.

Is that correct?

Update 2

This question is not about whether $f\colon z \mapsto z^{5/2}$ is holomorphic. I already stated a reason why it shouldn't. This question is about my inability to use the definition and the definition only to prove it is not holomorphic. Consequences of the definition include continuity, the Cauchy-Riemann equations, the local equivalence to Taylor series, Cauchy's integral theorem, etc. I am not asking 'how can I use those consequences of the definition to prove $f$ is not holomorphic?' My question is 'how do I use the definition to prove it?'.

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    $\begingroup$ The problem is that you can’t define $f$ continuous on any neighborhood of $0,$ even though you can make it continuous at $0.$ The function needs a “branch cut.” $\endgroup$ Nov 6, 2021 at 22:32
  • $\begingroup$ Also you have a Taylor series i.e. $\sum 0$ of radius $0$ centered in $0$, but you cannot extend it further on a little disk without having issues. $\endgroup$
    – zwim
    Nov 6, 2021 at 22:36
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    $\begingroup$ From the Wikipedia page: “ In mathematics, a holomorphic function is a complex-valued function of one or more complex variables that is complex differentiable in a neighbourhood of each point in a domain…” Neighborhood is the key word. Not just at a point. Implicitly, a holomorphic function is holomorphic on an open set, not just at a point. en.wikipedia.org/wiki/Holomorphic_function $\endgroup$ Nov 6, 2021 at 22:37
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    $\begingroup$ No, the update is not correct. The function is not continuous on a neighborhood of $0,$ so it is not holomorphic at $0.$ No idea of what you mean by Riemann sheet, but the definitions do not require an understanding of branches and Riemann manifolds, only that the definition is about a function differentiable on an open set, and there is no open neighborhood of $0$ where it is continuous, much less differentiable. $\endgroup$ Nov 6, 2021 at 22:55
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    $\begingroup$ @ThomasAndrews I meant Riemann surface. Anyway, the definition requires differentiability, which I understand and the existence of the limit of $(f(z_n)-f(z))/(z_n-z)$. Continuity might be a consequence of differentiability but that is not the definition itself (at least as I read it). $\endgroup$ Nov 6, 2021 at 23:02

3 Answers 3

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Emphatically, as in some comments, it is not possible to give a continuous function $f(z)$ on any nbd of $0$ such that $f(z)^2=z^5$. The issue doesn't even get to that of differentiability or not... Anticipation that $0$ is a "branch point" (whatever we decide that means) for any such $f(z)$ is a compelling indicator that there is simply no continuous function on any nbd of $0$ that could possibly be $z^{5/2}$.

The "negative" proof of this impossibility is fairly elementary, even without talking about "branches" or "sheets". For example, suppose there is a (convergent) power series $\sum_{n\ge 0} c_n\,z^n$ whose square is $z^5$... plain ol' algebra shows that there is no such power series, apart from convergence.

EDIT: in response to the edits to the question, I guess I'd ask what the "definition" of $z^{5/2}$ is that makes it continuous, much less differentiable (or not). Complex (or real) differentiability implies continuity... so if you can't give me a continuous function, I don't have to do anything at all to show it's not differentiable (in any elementary sense).

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  • $\begingroup$ Where is continuity required in the definition of holomorphism? $\endgroup$ Nov 6, 2021 at 23:12
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    $\begingroup$ @SolutionExists, being holomorphic (as "being differentiable" on $\mathbb R$) easily implies continuity. Is this your question? $\endgroup$ Nov 6, 2021 at 23:13
  • $\begingroup$ @SolutionExists If it was holomorphic at the origin it would also be continuous there, which is a contradiction. $\endgroup$
    – Gary
    Nov 6, 2021 at 23:19
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    $\begingroup$ @SolutionExists What do you get if you approach $0$ along $r e^{\pi i}$ and then along $r e^{-\pi i}$? You may use some rationalisation first to bring the differential quotient into an appropriate form. $\endgroup$
    – Gary
    Nov 6, 2021 at 23:24
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    $\begingroup$ @SolutionExists, the issue seems to be about specifying a $z^{5/2}$ which is continuous at $0$. It's not possible. $\endgroup$ Nov 6, 2021 at 23:29
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A way to look at things that may help is as follows: for every non zero complex number $z$ there are two complex numbers $y,w$ st $y^2=w^2=z^5$; now define a function $f$ in the plane by picking for $f(0)=0$ and for any other $z$ one of those two numbers by any method you want; for example you can choose some random way of assigning one of the numbers, or some clever way like when you pick a continuous argument in a slit plane and have a way to distinguish between the two numbers $y,w$ except on the slit (a smooth Jordan arc joining zero with infinity like a ray) itself.

Regardless of how you do it the function $f$ defined as above will be differentiable at zero and hence continuous there, though it can for example not be Lebesgue measurable on any plane domain. This shows that differentiability at one point is in general useless and what is important is differentiability in an open set which is what is equivalent to holomorphicity in the complex plane.

As others mentioned, however clever you are, you cannot choose a function $f$ as above to be continuous on an open set around zero, however small, though you can definitely choose it to be differentiable on small open sets around any other points and more generally on any slit plane where you exclude an arbitrary smooth Jordan arc joining zero with infinity

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  • $\begingroup$ If we consider the manifold of the graph of $f: z ↦ z^{5/2}$, the derivatives along any curve passing through the origin are well defined. In fact, $∇f$ is well defined at the origin. I used to think that complex derivatives were just directional derivatives with an extra step. That mental picture was reinforced by the fact that, in (at least two) textbooks, the definition of holomorphicity was given with one of the secant points being the point at which the tangent was being computed. $\endgroup$ Nov 9, 2021 at 19:04
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The problem I had was in using the definition of holomorphicity. The question was not about how the function fails continuity, the Cauchy-Riemann equations, the local equivalence to Taylor series, Cauchy's integral theorem, or any other statement that would follow from the definition.

The solution is rather simple: the definition of holomorphicity was wrong. While one can compute the derivative at a point using $\lim (f(x_0+h)- f(x_0))/h$, that result is not necessarily the derivative since one of the points was forced to be $x_0$.

For the example in the question, the following calculation shows that the derivative at the origin does not exist. $$ \lim_{n → ∞} \frac{e^{\frac{5 i π }{2 n}} - e^{ - \frac{5 i π }{2 n}}}{e^{\frac{i π }{n}} - e^{ - \frac{i π }{n}}} = \lim_{n → ∞} \left[ 2 \cos \left( \frac{3 π }{2 n} \right) + \frac 1 2 \sec \left( \frac π {2 n} \right) \right] = \frac 5 2 \neq 0. $$

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