Uniqueness of Analytic Continuation

Theorem

Let $U \subset V \subset \C$ be open subsets of the complex plane.

Let $V$ be connected.

Suppose:

$(1): \quad F_1, F_2$ are functions defined on $V$

$(2): \quad f$ is a function defined on $U$.

Let $F_1$ and $F_2$ be analytic continuations of $f$ to $V$.

Then $F_1 = F_2$.

Proof

Let $\map g z = \map {F_1} z - \map {F_2} z$.

Then:

$\forall z \in U: \map g z = 0$

Because Zeroes of Analytic Function are Isolated, and the zeroes of $g$ are not isolated, $g$ must be constant everywhere in its domain.

Since $\map g z = 0$ for some $z$, it follows that $\map g z = 0$ for all $z$.

Hence:

$\forall z \in V: \map {F_1} z - \map {F_2} z = 0$ and so $F_1 = F_2$.

$\blacksquare$