What is the analytic continuation of $$f(z) = \sum_{n=1}^{\infty} -\frac{z^n}n, \text{ where |z| < 1} $$

For real values of $z$, this function of course corresponds to the Taylor expansion of log(1-x). However, the complex logarithm function is multi-valued - so how do we choose an analytic continuation of $f(x)$ for $|z| > 1$, since the analytic continuation is supposed unique? Do we simply take the principial branch of the complex log function?

  • $\begingroup$ Yes, it is $Log (1-z)$ for $z \in \mathbb C \setminus (-\infty, 0]$ where Log is the principal branch. $\endgroup$ Nov 23, 2020 at 9:33
  • $\begingroup$ Take a look at this answer $\endgroup$
    – Jean Marie
    Nov 23, 2020 at 9:40
  • $\begingroup$ Why is it the principial branch? I looked at the linked answer but it doesn't really answer it. $\endgroup$ Nov 23, 2020 at 9:59
  • 1
    $\begingroup$ @KaviRamaMurthy You probably mean $z \in \mathbb C \setminus [1, \infty)$. The analytic continuation of $f$ to $\mathbb C \setminus [1, \infty)$ is unique, but there's nothing special about $[1, \infty)$, there's also a unique analytic continuation of $f$ to, say, $\mathbb C \setminus [1, 1 + i \infty)$. $\endgroup$
    – Maxim
    Nov 23, 2020 at 14:16
  • $\begingroup$ When it exists the analytic continuation of a power series at $a$ to a connected open containing $a$ is unique. Here the analytic continuation to $\Bbb{C}-1$ doesn't exist, it exists for any simply connected open $\subset \Bbb{C}-1$ containing $0$ and there it is unique. To understand the continuations of the continuations we need the concept of analytic continuation along a curve (the continuation of $\log z$ along the counterclockwise unit circle sends $\log z$ to $\log z+2i\pi$) $\endgroup$
    – reuns
    Nov 23, 2020 at 14:49

1 Answer 1


Analytic continuation is not as universally applicable as we might sometimes want. The identity theorem says that if $f$ and $g$ are analytic on a open connected domain $D$ and agree on an open subset of $D$ (or even on a set of points that has am accumulation point in $D$) then $f=g$ throughout $D$. But notice that $f$ and $g$ must first both be analytic on $D$. You cannot always guarantee that you can extend $f$ from a smaller to larger domain unless you first know such an extension will work.

An example is as follows. We know one version of $\log z$ can be defined that is analytic on the domain \begin{align} D_1 = \{ z: 1 < |z| < 3, |\arg(z)| > \pi/8\}. \end{align}
Using a power series centred at $z = 2$ we can create a second function analytic in the disc $D_2 = \{z: |z-2| < 2\}$ and which agrees with our $\log z$ on the part of $D_1 \cap D_2$ for which $\Im z > 0$. But this will not then agree with our $\log z$ on the intersection $D_1 \cap D_2$ where $\Im z < 0$. That is because there is no analytic function on a domain that circles the origin and at the same time matches our $\log z$ throughout $D_1$. In this example the identity theorem cannot be used as such an extension does not exist. Picture of two domains <span class=$D_1$ and $D_2$" />

Turning to your example, you can extend your power series into a larger domain, but you may end up with different branches depending on how your extension progresses around the singularity, which in your case is at $z=1$. But as soon as you try to make your extension circle back to the first domain, the identity theorem ceased to apply.


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