Identity theorem

If two holomorphic functions f and g on a domain D agree on a set S which has an accumulation point c in D, then f = g on a disk in ${\displaystyle D}$ centered at ${\displaystyle c}$.

To prove this, it is enough to show that ${\displaystyle f^{(n)}(c)=g^{(n)}(c)}$ for all ${\displaystyle n\geq 0}$.

If this is not the case, let m be the smallest nonnegative integer with ${\displaystyle f^{(m)}(c)\neq g^{(m)}(c)}$. By holomorphy, we have the following Taylor series representation in some open neighborhood U of c:

${\displaystyle {\begin{aligned}(f-g)(z)&{}=(z-c)^{m}\cdot \left[{\frac {(f-g)^{(m)}(c)}{m!}}+{\frac {(z-c)\cdot (f-g)^{(m+1)}(c)}{(m+1)!}}+\cdots \right]\\[6pt]&{}=(z-c)^{m}\cdot h(z)\end{aligned}}}$

By continuity, h is non-zero in some small open disk B around c. But then f − g ≠ 0 on the punctured set B − {c}. This contradicts the assumption that c is an accumulation point of {f = g}.

This lemma shows that for a complex number a, the fiber f⁻¹(a) is a discrete (and therefore countable) set, unless f ≡ a.

Define the set on which ${\displaystyle f}$ and ${\displaystyle g}$ have the same Taylor expansion:

${\displaystyle S=\{z\in D\mid f^{(k)}(z)=g^{(k)}(z){\text{ for all }}k\geq 0\}=\bigcap _{k=0}^{\infty }\{z\in D\mid (f^{(k)}-g^{(k)})(z)=0\}.}$

We'll show ${\displaystyle S}$ is nonempty, open, and closed. Then by connectedness of ${\displaystyle D}$, ${\displaystyle S}$ must be all of ${\displaystyle D}$, which implies ${\displaystyle f=g}$ on ${\displaystyle S=D}$.

By the lemma, ${\displaystyle f=g}$ in a disk centered at ${\displaystyle c}$ in ${\displaystyle D}$, they have the same Taylor series at ${\displaystyle c}$, so ${\displaystyle c\in S}$, ${\displaystyle S}$ is nonempty.

As ${\displaystyle f}$ and ${\displaystyle g}$ are holomorphic on ${\displaystyle D}$, ${\displaystyle \forall w\in S}$, the Taylor series of ${\displaystyle f}$ and ${\displaystyle g}$ at ${\displaystyle w}$ have non-zero radius of convergence. Therefore, the open disk ${\displaystyle B_{r}(w)}$ also lies in S for some r. So S is open.

By holomorphy of ${\displaystyle f}$ and ${\displaystyle g}$, they have holomorphic derivatives, so all ${\displaystyle f^{(n)},g^{(n)}}$ are continuous. This means that ${\displaystyle \{z\in D\mid (f^{(k)}-g^{(k)})(z)=0\}}$ is closed for all ${\displaystyle k}$. ${\displaystyle S}$ is an intersection of closed sets, so it's closed.

• Analytic continuation

• Ablowitz, Mark J.; Fokas A. S. (1997). Complex variables: Introduction and applications. Cambridge, UK: Cambridge University Press. p. 122. ISBN 0-521-48058-2.