radius of convergence of Taylor series, function with branch cuts

Question

Let $f(z) $ being the analytic continuation of some holomorphic function, having many branch points and isolated singularities at $\beta_1,\beta_2,\ldots,\beta_n,\ldots$

is the radius of convergence of its Taylor series at $a$ still given by :

$$R = \min_{n} |a - \beta_n| $$
even if as $f(z)$ is defined, there is one (or more) branch cut traversing the disk $|z-a| < R$ ?

(I'm not asking for particular cases such as $ f(z) = \log(z)$, but how to prove it in the general case where we don't know anything on $f$ except what I wrote)

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EDIT : the answer is no, an example is given by Mercio : by considering $g(z) = \log(z)$ analytic on $|z-1| < 1$ such that $g(1) = 0$, and following it by analytic continuation on a spiral of increasing radius around $0$. After a full rotation we arrive say at $z= 7$ but on the branch of $ \log(z)$ for which $\log(7) = 2 i \pi + \ln 7$, and on that branch there is no zero at $z=1$, hence the Taylor series of $$f(z) = \frac{1}{g(z)}$$ at $z= 7$ has radius of convergence $7$ while $$\min_n |7 - \beta_n| = 6$$

Answer 1

What you call $f$ is not a function but an "analytisches Gebilde" (analytic something) resulting from the analytic continuation of function elements $(f_\iota,U_\iota)_{\iota\in I}$, each of them defined in some open set $U_\iota\subset{\mathbb C}$. Note that one and the same point $z_0\in{\mathbb C}$ will in general be covered by several $U_\iota\,$, and the corresponding $f_\iota$ will have different values at $z_0$. There are no branch cuts coming with such an $f$. Branch cuts are an accessory device brought in by the engineering analyst in order to facilitate and standardize the handling of a given particular $f$.

Assume now that $f_\iota:\>U_\iota\to{\mathbb C}$ is a function element of $f$. This $f_\iota$ is a standard holomorphic function on the domain $U_\iota$. For $a\in U_\iota$ the Taylor series of $f_\iota$ at $a$ has the form $$f_\iota(z)=\sum_{k\geq0}c_k(z-a)^k\ ,\tag{1}$$ and its radius of convergence is the sup of all $\rho>0$ such that $f_\iota$ is analytic in the disk $D_\rho(a)$. It may very well be that some of the points $\beta_k$ are lying in the domain of convergence of $(1)$ because such points are singularities of other branches of $f$ near $a$.

Answer 2

Branch cuts are an artefact of people who insist on wanting to graph a function on the complex plane when really they shouldn't.

Branch cuts are ugly.

A branch cut is not an essential property of a complex function.
How can you be sure that when you talk about "$\log(z)$", your interlocutor is using the same branch cut as you ? You can't.

The radius of convergence is a local property of a function, it is the upper bound on the set of radii for which you can analytically continue the germ of your function on an open ball centered around your point. But if you make bad enough branch cuts, the poles, singularities, and branch points that you see may not be the ones relevant.

It turns out that when I have to choose a branch cut for $\log(z)$, noone is stopping me from choosing the curve $\gamma(t) = te^{it}$ for $t\ge 0$.

When I plot the function $1/\log(z)$ (with the branch of $\log$ where $\log(1)=0$), and look at the point $z=7$, well I have a branch point at $0$ and a pole at $1$, so the radius of convergence should be $6$ right ? Wrong ! the radius is $7$.

In fact you can do even worse, for example if I take the branch where $\log(1) = -2i\pi$ and plot the branch of $1/\log(z)$ with my branch cut, the radius of convergence at $z=7$ looks like it should be $7$ because I only see a branch point at $0$ and no pole, but in this case the radius is actually $6$ !

So really, stop thinking in terms of branch cuts.

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You should think of the graph of "$\log(z)$" as the subset $G = \{(\exp w,w) ; w \in \Bbb C \} \subset \Bbb C^2$. This subset looks like a graph of a holomorphic function if you don't look around too much, but when you look at it globally, it turns out it isn't.

But, this subset still allows you to talk about the radius of convergence of the function it looks like locally at a point $(z,y)$, as the least upper bound of the set of radii $R$ for which the connected component of $(z,y)$ in $G \cap B(z,R) \times \Bbb C$ is the graph of a holomorphic map $B(z,R) \to \Bbb C$.

And even then, sometimes the continuation of the graph is not a nice enough object to study the analytic continuation of a function, that's why we eventually need to use general complex manifolds $M$ with canonical maps $z : M \to \Bbb C$. Then a function on $M$ is a holomorphic map $f : M \to \Bbb C$ and you can "plot" $f$ by plotting $(z(m),f(m))$. In the case of $\log(z)$, $M$ is just $\Bbb C$, $f$ is the identity, and $z$ is the exponential map.