Uniqueness of Analytic Continuation

Theorem
Let $$U \subset V \subset \C \ $$ be open subsets of the complex plane, and suppose $$F_1, F_2 \ $$ are functions defined on $$V \ $$, and $$f \ $$ is a function defined on $$U \ $$.

If $$F_1 \ $$ and $$F_2 \ $$ are analytic continuations of $$f \ $$ to $$V \ $$, and $$V \ $$ is connected, then $$F_1 = F_2 \ $$.

Proof
Let $$g(z)=F_1(z)-F_2(z) \ $$. Then for $$z \in U, g(z) =0 \ $$. 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 $$g(z) = 0 \ $$ for some $$z, g(z) = 0 \ $$ for all $$z \ $$. Hence $$F_1(z)-F_2(z) =0 \forall z \Longrightarrow F_1=F_2 \ $$.