In Knopp's theory of functions part 1, the following fact about analytic continuation is mentioned without proof. And I am looking for a simple proof.
Let $$f(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + \dots$$ be a power series around $z_0$ with radius of convergence $R$ where $0 < R < \infty.$
Let $z_1$ be a point in the region of convergence with $ 0 < |z_1 – z_0| < R.$
Assume that $f$ can be expanded in a power series around $z_1$ with a radius of convergence exactly equal to $R_1 = R – |z_0 – z_1|$.
Clearly, $C$, the circle with center $z_0$ with radius $R$ and, $C_1$ the circle with its center at $z_1$ with radius $R_1$ intersect at exactly one point, say $z_2$ and $C_1$ lies within $C$.
Knopp mentions that $z_2$ is a singular point of $f$ in the following sense : given any neighborhood of $z_2$,say $U = B(z_2,r)$, there is no analytic function, $g$ defined on $U$ whose values coincide with the values of $f$ on $U \cap B(z_0,R)$.
My observations :
To prove this, it is sufficient to show that given any $r > 0$ there exists a $\delta > 0$ such that $B(z_1,R_1+\delta) \subseteq B(z_0,R) \bigcup B(z_2,r) \tag{*} \label{e:1}. $
To see this notice that if an analytic function $g$ exists on $B(z_2,r)$ for some $r > 0$, such that the values of g coincide with the values of $f$ on $B(z_0,R) \bigcap B(z_2,r)$ then $g$ is an analytical extension of $f$ on $B(z_0,R) \bigcup B(z_2,r)$ and moreover assuming $\eqref{e:1}$ it means that $f$ can be extended to an analytical function on $B(z_1,R_1+\delta)$. However this means the radius of convergence of the power series determined by $f$ at $z_1$ must be strictly larger than $R_1$ which is a contradiction.
So it remains to prove $\eqref{e:1}$ which is a purely geometric problem. A visual representation suggests that if $\delta$ is chosen to be less than the length of a segment joining $z_1$ and $A$ where $A$ is a point of intersection of the circle at $z_2$ with radius $r$ and $C$ the circle with center at $z_0$ with radius $R$ we should be good. But I don't see an easy proof.
Best Answer
The simplest proof, if one knows a little topology, is to note that $\overline{B(z_1,R_1)}$ is a compact subset of the open set $$U = B(z_0,R) \cup B(z_2,r)\,.$$
The continuous function $z \mapsto \operatorname{dist}(z,\mathbb{C}\setminus U)$ is strictly positive on $\overline{B(z_1,R_1)}$, hence it attains a strictly positive minimum $\eta$ there, and any $\delta \in (0,\eta]$ works.
This does not give an explicit expression for a viable $\delta$, though, unless one works through the geometry (which would make the appeal to compactness superfluous, but is more work).