limit point implies uniformly zero

something pretty cool in complex is that if you have these two things for a domain $\Omega$ and a holomorphic function $f$:

a sequence of distinct points in $\Omega$ $a_1, a_2, \cdots$ s.t. $\lim_{n \to \infty} a_n = a$
$f(a_i)=0$ for all $i \in \N$

then this literally means $f$ is 100% zero everywhere in $\Omega$ (and i guess $\C$ by extension)

zero in disc

how the proof first goes is that you show that $f$ is always zero on some tiny disc around $a$

this part is proven by contradiction, so say

\[f(z)=\sum_{n=N}^\infty a_n(z-a)^n\]

where $a_N \ne 0$.

actually given this we can factor out the lowest power and write

\[f(z)=(z-a)^N g(z)\]

where $g(a) \ne 0$ and by continuity there should be some radius around $a$ where $g(z) \ne 0$ as well

thing is, since we have a convergent sequence of zeros for any $R > 0$ $\exists z \in B_R(a): z \ne a \land f(z)=0$.

in our rewritten form, this gives $(z-a)^N g(z)=0$. the first coefficient can't be 0, since $z \ne a$, so $g(z)=0$. this contradicts our earlier statement that $g(z)$ had some tiny area around $a$ where it was always nonzero.

so yeah, $f$ has to be constantly 0 in some ball

zero EVERYWHERE!

now we extend the places where $f$ is zero to literally everywhere

it suffices to show that the set here:

\[A=\{z \in \Omega \mid \exists r: f(z')=0\ \forall z' \in B_r(z)\}\]

is both closed and open. this is because $\Omega$ is a domain and therefore connected, and the only subsets of it that are both closed and open inside it are the null set and $\Omega$ itself.

this set is defo nonempty since $a \in A$, so we only have to show clopenness

open

for openness, for any $z_0 \in A$ we can take its radius $r$ and say that any $z \in B_{r/3}(z_0)$ is also in $A$, since they're locally zero in a radius of $\frac{r}{3}$ as well.

these choices of points and radii work since $z \in B_{r/3}(z_0) \implies B_{r/3}(z) \subseteq B_r(z_0)$. if we're zero everywhere in that larger ball, then surely we're zero everywhere in the smaller ball too, right?

closed

i'll show the following implication for a point $z$. all it's really saying is that $\overline{A} \subseteq A$, so $A$ is closed.

\[\forall r > 0\ B_r(z) \cap A \ne \varnothing \implies z \in A\]

this actually follows pretty directly from continuity, since the premise implies a sequence of points converging to $z$ that all have a value of $0$. by continuity of $f$, $f(z)=0$ as well and so $z \in A$

but yeah, that's clopenness and the whole thing proved i think $\square$