Uniqueness of Analytic Continuation

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Theorem

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

Let $V$ be connected.


Suppose:


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 $F_1 = F_2$.

$\blacksquare$