Given the two functions: $$f_1(z) = \sum_{n=1}^{\infty}\frac{z^n}{n}\ \ \text{and}\ \ f_2(z) = i\pi\ +\ \sum_{n=1}^{\infty}(-1)^n\frac{(z-2)^n}{n}$$ I have to show that they are analytic continuations of one another.
Is this even possible since the two series have no common domain of convergence? How should I go about doing this?
Any help would be greatly appreciated.
The region of absolute convergence of $f_1$ is $\|z\|< 1$ and the region of absolute convergence of $f_2$ is $\|z-2\|<1$, so we are not that lucky, since these regions do not overlap over an open set. On the other hand, we may try to show that $f_1,f_2$ are restriction of the same holomorphic function $f_3$ defined over the following awkward set: $$S=\{z\in\mathbb{C}:\|z-1\|<2\}\setminus\{z\in\mathbb{C}:z=1-it,t\geq 0\},$$ containing both the previous domains of absolute convergence. The formal derivative of $f_1$ is $\frac{1}{1-z}$ and the formal derivative of $f_2$ is $\frac{1}{1-z}$: at least the hunt makes sense. The formal primitive of $\frac{1}{1-z}$ is $-\log(1-z)$ and such function has a holomorphic determination over $S$, such that $-\log(1-2)=\pi i=f_2(2)$ and $-\log(1-0)=0=f_1(0)$. I bet you can finish from here.