We have shown that a holomorphic map \(f: G\to \mathbb{C}\) to be expressed as a power series, which bears a certain similarity to polynomials, and a feature of polynomials are that if \(a\) is a root, or zero, for a polynomial \(p\), we can factor \(p\) such that \(p(z)=(z-a)^n q(z)\) where \(q\) is another polynomial with the property that \(q(a)\neq 0\). Now, does this similarity with polynomials extend to factorization? In fact it does as we shall see.

Let \(f: G\to \mathbb{C}\) be a holomorphic map that is not identically zero, with \(G\subseteq \mathbb{C}\) a domain and \(f(a)=0\). It is our claim that there exists a smallest natural number \(n\) such that \(f^{(n)}(a)\neq 0\). So suppose that there are no such \(n\), such that \(f^{(k)}(a)=0\) for all \(k\in\mathbb{N}\). Let \(B_\rho(a)\) be the largest open ball with center \(a\) contained in \(G\), since we have that \[f(z)=\sum^\infty_{k=0}\frac{f^{(k)}(a)}{k!}(z-a)^k\] we then have that \(f\) is identically zero on \(B_\rho(a)\). Fix a point \(z_0\in G\) and let \(\gamma : [0,1]\to G\) be a continuous curve from \(a\) to \(z_0\). By the paving lemma there is a finite partition \(0=t_1 < t_2 <\cdots <t_m=1\) and an \(r>0\) such that \(B_r(\gamma(t_k))\subseteq G\) for all \(k\) and \(\gamma([t_{k-1},t_k])\subseteq B_r(\gamma(t_k))\). Note that \(B_r(\gamma(t_1))=B_r(a)\subseteq B_\rho(a)\) so \(f\) is identically zero on \(B_r(\gamma(t_1))\), but since \(\gamma([t_1,t_2])\subseteq B_r(\gamma(t_1))\) we must have that \(f\) is identically zero on \(B_r(\gamma(t_2))\), and so on finitely many times untill we reach \(\gamma(t_m)\) and conclude that \(f\) is identically zero on \(B_r(\gamma(t_m))=B_r(z_0)\) and since \(z_0\) was chosen to be arbitrary we must conclude that \(f\) is identically zero on all of \(G\). A contradiction.

Now, let \(n\) be the smallest natural number such that \(f^{(n)}(a)\neq 0\), then we must have that \(f^{(k)}(a)=0\) for \(k < n\). We then get, for \(z\in B_\rho(a)\): \[\begin{split} f(z) &=\sum^\infty_{k=0}\frac{f^{(k)}(a)}{k!}(a-z)^k \\ &= \sum^\infty_{k=n}\frac{f^{(k)}(a)}{k!}(a-z)^k \\ &= \sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{n+k} \\&=(z-a)^n \sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{k}, \end{split}\] now, let \(\tilde{f}(z)=\sum^\infty_{k=0}\frac{f^{(n+k)}(a)}{(n+k)!}(a-z)^{k}\) and note that \(\tilde{f}\) is non-zero and holomorphic on \(B_\rho(a)\). We then define a map \(g\) given by \[g(z)=\begin{cases} \tilde{f}(z), & z\in B_\rho(a) \\ \frac{f(z)}{(z-a)^n}, & z\in G\setminus \{a\}\end{cases}\] and note that \[f(z)=(z-a)^n g(z),\] showing the existance of a factorization with our desired properties. Showing that this representation is unique is left as an exercise 😉

## References

*Complex analysis*. Department of Mathematical Sciences, University of Copenhagen.