Uniqueness of Analytic Continuation

Theorem

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

Let $V$ be connected.

Suppose:

• $F_1, F_2$ are functions defined on $V$;
• $f$ 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 $g \left({z}\right) = F_1 \left({z}\right) - F_2 \left({z}\right)$.

Then:

$\forall z \in U: g \left({z}\right) = 0$

Since the zeroes of non-constant analytic functions are isolated, and the zeroes of $g$ are not isolated, $g$ must be constant everywhere in its domain.

Since $g \left({z}\right) = 0$ for some $z$, it follows that $g \left({z}\right) = 0$ for all $z$.

Hence:

$\forall z \in V: F_1 \left({z}\right) - F_2 \left({z}\right) = 0$ and so $F_1 = F_2$.

$\blacksquare$