# Analytic continuation of a logarithm function

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?

• Yes, it is $Log (1-z)$ for $z \in \mathbb C \setminus (-\infty, 0]$ where Log is the principal branch. Nov 23, 2020 at 9:33
• Take a look at this answer Nov 23, 2020 at 9:40
• Why is it the principial branch? I looked at the linked answer but it doesn't really answer it. Nov 23, 2020 at 9:59
• @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, $\mathbb C \setminus [1, 1 + i \infty)$. Nov 23, 2020 at 14:16
• 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 $\log z+2i\pi$) Nov 23, 2020 at 14:49

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. $D_1$ 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.