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)

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 $$\min_n |7 - \beta_n| = 6$$

  • $\begingroup$ Yes. See e.g. math.stackexchange.com/questions/574816/… for this exact question for the spesific case of $f(z) = \log(z)$. See also math.stackexchange.com/questions/1556805/… for the case of $f(z) = \sqrt{1-z^2}$. $\endgroup$
    – Winther
    May 21, 2016 at 1:39
  • $\begingroup$ I don't see that you need to make any modification to the standard proof that the series converges (e.g. this). All that is required is that $f$ is analytic inside a disc of radius $R$ about $z_0$ and that is guaranteed by your conditions. The fact that a branch cut passes though does not change the convergence properties of the series as long as the branch-point is outside this disc. $\endgroup$
    – Winther
    May 21, 2016 at 2:26
  • $\begingroup$ @Winther : $f(z)$ is not holomorphic inside the disc if there is a branch cut traversing it... so you have to prove first that you can move the branch cuts out of the disk. once this is done, you get $f_2(z)$ which is the analytic continuation of $f(z)$ and that is holomorphic on the disk and whose Taylor series at $a$ has a ROC obviously $\ge R$, and since the Taylor series of $f(z)$ is $\sim$ the same (have to prove this too, i.e. that moving the branch cuts didn't change $f(z)$ on the disk), it'll have a ROC $\ge R$ too. $\endgroup$
    – reuns
    May 21, 2016 at 2:57
  • $\begingroup$ But a branch cut is arbitrary in the sense that you can place it almost wherever you want. It does not have to be a straight line either and the only restriction is that all branch points have to lie on the line. Just choose the cut such that it does not intersect the circle. $\endgroup$
    – Winther
    May 21, 2016 at 3:00
  • $\begingroup$ what's a branch cut ? $\endgroup$
    – mercio
    May 31, 2016 at 8:25

2 Answers 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.

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.

  • $\begingroup$ now you example of $1/\log(z)$ is indeed a good example, you had to write only this (I just understood, your message is way too long) : by choosing the branch cuts of $g(z) = \log(z)$ correctly, $f(z) = 1/g(z)$ has a zero at $z=1$, but at $z=4$ it is locally the branch of $\log(z)$ for which $\log(1) = 2 i \pi$, and consequently its Taylor series at $z = 4$ has radius of convergence $4$, not $3$ $\endgroup$
    – reuns
    May 31, 2016 at 9:16
  • $\begingroup$ so you agree that this example should be teached in complex analysis courses ? that the radius of convergence of the Taylor series can be complicated to know for those kind of functions ? (and yes I know how to see those functions as being analytic in Riemann surfaces as in the drawing of $\log(z)$ there ) $\endgroup$
    – reuns
    May 31, 2016 at 9:24
  • $\begingroup$ I think it should really be emphasized that branch cuts are completely arbitrary and it's no use to define a function $\log$ (or Lambert W or etc) if you don't ask yourself where you want it to be defined and how. In fact I didn't need to pick a bad branch cut for the 2nd example, just hiding a pole just behind the normal branch cut is enough to get a "lying" picture. $\endgroup$
    – mercio
    May 31, 2016 at 9:30
  • $\begingroup$ I don't get what you mean. the branch cuts arise when projecting in some way an function analytic/meromorphic on a Remann surface to the complex plane !? and I could not define my quesiton in an other way (I tried before with : can the singularities move/disappear/appear when rotating around a branch point, and I got even less comments/interest) $\endgroup$
    – reuns
    May 31, 2016 at 9:34
  • $\begingroup$ now do you know what happens with functions of several complex variables ? it seemed when searching it is even worse in this case. $\endgroup$
    – reuns
    May 31, 2016 at 9:38

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 $a$.

  • $\begingroup$ I start with a function $f(z)$ which is the analytic continuation of $g(z) : B_r(z_0) \to \mathbb{C}$. this function has a some singularities, and on some lines/curves it is discontinuous : it has branch cuts. now I go at some point $a$ close to one of those lines (point where it is analytic) and I compute its Taylor series. $\endgroup$
    – reuns
    May 31, 2016 at 9:07
  • $\begingroup$ my question was indeed : can some singularities move/disappear/appear when rotating around a branch point. $\endgroup$
    – reuns
    May 31, 2016 at 9:28

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