# Complex Analysis: #10 Analytic Continuation

_{0}∈ G. The function f is equal to the function defined by the power series in the largest possible open disc around z

_{0}which is contained in G. But of course G is not, in general itself an open disc. Therefore there might be parts of G where f is not given by this power series centered on z

_{0}. Or, (thinking about the logarithm function) we might have the situation that G could be expanded to a larger region G

_{*}, with G ⊂ G

_{*}where the function could be defined. But then perhaps there might be different ways of “continuing” this definition of f from G to G

_{*}.

Therefore let us consider a chain of open discs (B

_{1}, . . ., B

_{n}) say, with

B

_{j}= {z ∈ ℂ : |z − p_{j}| < r_{j}}for a chain of points p

_{j}which are the centers of the discs, and numbers r

_{j}> 0, which are the radii in each case. We assume that it is a connected chain in the sense that B

_{j}∩ B

_{j+1}≠ ∅, in each case. For each B

_{j}let us assume that an analytic function f

_{j}: B

_{j}→ ℂ exists, such that in the region of overlap, we have f

_{j}(z) = f

_{j+1}(z) for z ∈ B

_{j}∩ B

_{j+1}.

**Deﬁnition 8**

The functions f

_{j}here are called

*function elements*, and if we have f

_{j}(z) = f

_{j+1}(z) for z ∈ B

_{j}∩ B

_{j+1}for all j = 1, . . . , n − 1, then we have an analytic continuation of the function elements through the chain of open discs. Or, if we consider the ordering of the discs, we can say that the ﬁnal function f

_{n}is obtained by analytic continuation of the initial function f

_{1}through the chain of discs.

Theorem 17 shows that if, say f

_{1}is given in B

_{1}, and there exists a chain of open discs allowing some analytic continuation, then this continuation is

*unique*.

**Theorem 21**Let (B

_{1}, . . ., B

_{n}) be a chain of open discs with B

_{j}∩ B

_{j+1}≠ ∅, for j = 1, . . . , n − 1. Let f

_{1}: B

_{1}→ ℂ be some given analytic function. Then there exists an analytic continuation of f

_{1}throughout the chain [That is to say, there exists a set of function elements forming an analytic continuation, such that the ﬁrst element in the chain of function elements is f

_{1}.] if and only if there also exists an analytic continuation of f

_{1}' : B

_{1}→ ℂ (the derivative of f

_{1}) throughout the chain.

*Proof*

“⇒” is trivial. (Just take f

_{j}', the derivative of f

_{j}, for each j.)

As far as “⇐” is concerned, we are assuming that there is an analytic continuation of the function . To avoid confusion, let us call this function g

_{j}: B

_{1}→ ℂ. i.e. g

_{1}(z) = f

_{1}'(z) for all z ∈ B

_{j}. The assumption is that for each j there is an analytic function g

_{j}: B

_{j}→ ℂ, providing an analytic continuation, starting with g

_{1}. We now use induction on the number n. For n = 1 there is nothing to prove. So let n > 1, and assume that we have an analytic continuation of g along the chain of open discs from B

_{1}to B

_{n}

_{−1}, giving g

_{j}: B

_{j}→ ℂ such that g

_{j}: f

_{j}' for each j < n. According to theorem 3, g

_{n}has an antiderivative, G

_{n}: B

_{n}→ ℂ, with G

_{n}' : g

_{n}. Now in the region B

_{n}

_{−1}

_{}∩ B

_{n}we have g

_{n}

_{−1}= g

_{n}. That is, f

_{n}

_{−1}' = G

_{n}', or f

_{n}

_{−1}' − G

_{n}' = 0. Thus f

_{n}

_{−1}− G

_{n}= k say, where k is a constant number. But then the function G

_{n}+ k is also an antiderivative to g

_{n}, and we can take f

_{n}= G

_{n}+ k.

One way to think about these chains of discs is to imagine that they are associated with a path, namely a path starting at p

_{1}then following a straight line to p

_{2}, then a straight line to p

_{3}, and so forth, ﬁnally ending at p

_{n}. So each straight segment, say from p

_{j}to p

_{j+1}is contained in the union of the two discs B

_{j}∪ B

_{j+1}.

Let’s generalize this idea in the following way. Let γ : [t

_{0}, t

_{1}] → ℂ be a continuous path. (It doesn’t have to be differentiable here.) Let t

_{0}= τ

_{0}< τ

_{1}< · · · < τ

_{n}= t

_{1}be a partition of the interval [t

_{0}, t

_{1}]. Assume that we have a corresponding set of open discs B

_{j}= {z ∈ ℂ : |z − γ(τ

_{j})| < r

_{j}}(where r

_{j}> 0) such that γ(t) ∈ B

_{j}∪ B

_{j+1}, for t ∈ [τ

_{j}, τ

_{j+1}], for all relevant j. Then we will say that we have a chain of discs along the path γ. Furthermore, if we have a sequence of analytic function elements giving an analytic continuation along this chain, then we will say that the function is analytically continued along the path. The next theorem shows that this is a property of the path, independent of the choice of discs along the path.

**Theorem 22**Let γ : [t

_{0}, t

_{1}] → ℂ be continuous and let B = {z ∈ ℂ : |z − γ(t

_{0}) < r} and B

_{*}= {z ∈ ℂ : |z − γ(t

_{1})| < r

_{*}} (with both r, r

_{*}> 0) be open discs centered on γ(t

_{0}) and γ(t

_{1}), respectively. Let t

_{0}= τ

_{0}< τ

_{1}< · · · < τ

_{n}= t

_{1}and t

_{0}= ρ

_{0}< ρ

_{1}< · · · < ρ

_{m}= t

_{1}be two different partitions of the interval [t

_{0}, t

_{1}], giving two different chains of discs along the path, satisfying the conditions listed above, where the ﬁrst and last discs are B and B

_{*}. Assume that f : B → ℂ is an analytic function at the ﬁrst disc and it has an analytic continuation with respect to the ﬁrst chain of discs, ﬁnally giving the analytic function g : B

_{*}→ ℂ. Then f also has an analytic continuation with respect to the second chain of discs, and it also gives the same function g : B

_{*}→ ℂ.

*Proof*

If the whole path γ is completely contained within the disc B then, using theorem 9 we see that the analytic continuation is simply given by the power series representing the function.

Therefore let M ⊂ [t

_{0}, t

_{1}] be defined to be the set of t

_{*}∈ [t

_{0}, t

_{1}] such that the theorem is true for the interval [t

_{0}, t

_{*}] (with respect to this path γ). But if t ∈ M, then since γ is continuous, and since the analytic function which has been continued out to the point γ(t

_{*}) has a power series representation in a neighborhood of γ(t

_{*}), we must have some ∈ > 0 such that {t ∈ [t

_{0}, t

_{1}] ; |t − t

_{*}| < ∈ } ⊂ M. Therefore M is an open subset of [t

_{0}, t

_{1}]. If [t

_{0}, t

_{1}] \ M ≠ ∅, then the same argument shows that M is closed in [t

_{0}, t

_{1}]. Yet the interval [t

_{0}, t

_{1}] is connected. Therefore M = [t

_{0}, t

_{1}].

All of these thoughts allow us to perform path integrals along continuous, but not necessarily differentiable paths. To see this, take γ to be some continuous path, allowing an analytic continuation from an open disc centered on the starting point of the path. The discs in the ﬁnite chain of open discs describing the analytic continuation have centers at various points along the path γ. But now take γ to be the piecewise linear path connecting those centers. This gives a path integral, namely ∫

_{γ}f(z)dz, where f consists of the function elements along the path. We can now simply define ∫

_{γ}f(z)dz to be this integral along γ. Theorem 22 then shows that the path integral for γ is well defined.

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