I am reading Conway's complex analysis textbook, vol I.
I have this question which I want to see if I have the right idea, and if not correct it.
The collection $\{D_0,D_1,...,D_n \}$ of open disks is called a chain of disks if $D_j \cap D_{j-1} \neq \emptyset$ for $1\leq j \leq n$.
If $\{(f_j,D_j): 0\leq j \leq n\}$ is a collection of function elements such that $\{D_0,D_1,...,D_n \}$ is a chain of disks and $f_j(z)=f_{j-1}(z)$ for $z\in D_j \cap D_{j-1}$, $1\leq j\leq n$; then $\{(f_j,D_j)\}$ is called an analytic continuation along a chain of disks.
(a) Let $\{(f_j,D_j)\}$ be such an analytic continuation along a chain of disks, and let $a$ and $b$ be the centres of the disks $D_0$ and $D_n$ resepcetively. Show there's a path, $\gamma$ from $a$ to $b$ and an analytic continuation $\{(g_t,B_t)\}_{t \in [0,1]}\}$ along $\gamma$, s.t the curve is contained in the union of all the disks, and $[f_0]_a=[g_0]_a$, $[f_n]_b=[g_1]_b$.
(b)The converse of (a).
Now for (a), I think to start at the centre $a$, and draw a straight line from $a$ to a point on $\partial D_1$ which is the closest to a, i.e take the infimum of the distance from a to the boundary of $D_1$, and for the point of intersection of this line with the boundary of $D_1$ call $a_1$, and from this point again draw such a line with the boundary of $D_2$, and proceed this way, thus we get segments of lines which together make a continuous function, I am not sure if it's possible to take these lines as phase lines which are tangent to the curve itself (in order for the path to be smooth).
For $g_t$, I just define $f_j=g_{\frac{j}{n}}$, and $g_0=f_0$, and for any other values $(j-1)/n < t \leq j/n$ $f_j=g_{\frac{j}{n}}=g_t$.
I haven't yet thought of (b), yet.