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?
Best Answer
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}
$D_1$ and $D_2$" />
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.
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.