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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.

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You have carefully defined what the analytic continuation along a chain of disks means. How is the analytic continuation $\{(g_t,B_t)\}_{t\in[0,1]}$ defined? – robjohn Dec 2 '11 at 20:50
You should avoid points on $\partial D_i$, as the corresponding $f_i$ is not defined there. Choose points $z_i\in D_{i-1}\cap D_i$ as intermediate marks instead. – You are not required to produce a smooth path; a continuous path suffices. – Christian Blatter Dec 2 '11 at 21:11
@ChristianBlatter, yes I understood it after I posted my question, that it's better to choose points in the intersection. – MathematicalPhysicist Dec 3 '11 at 9:27

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