Factorizing holomorphic functions

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 😉


Berg, C. (2016). Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen.

Pitfalls of open maps

Suppose you have an open map \(p\) between topological spaces, and if you have a subet \(A\) of \(p\)’s domain such that \(p(A)\) is open. Can you then conclude that \(A\) is open? Nope! Consider the following spaces \(X=\{x_1,x_2\}\) and \(Y=\{y_1,y_2\}\) with topologies \(\tau_X=\{\varnothing, X, \{x_1\}\}\) and \(\tau_Y=\{\varnothing,Y,\{y_1\}\}\), respectively and let \(p: X\times Y\to X\) be the projection onto its first fator. This is an open map. If we consider \(A=X\times\{y_2\}\) we see that \(A\) is not open in \(X\times Y\), but we have that \(p(A)=p(X\times\{y_2\})= X\) which is trivially open in \(X\).

Is this quotient space connected?

I came across this little problem recently: If \(X\) is a topological space with exactly two components, and given an equivalence relation \(\sim\) what can we say about its quotient space \(X/{\sim}\)? It turns out that \(X/{\sim}\) is connected if and only if there exists \(x,y\in X\) where \(x\) and \(y\) are in separate components, such that \(x\sim y\).

Suppose first that there exists \(x,y\in X\) such that \(x\sim y\). Let \(C_1\) and \(C_2\) be the two components of \(X\) and let \(p: X \to X/{\sim}\) be the natural projection. Since \(p\) is a quotient map it is surely continuous and since the image of a connected space under a continuous function is connected we have have that, say \(p(C_1)\) is connected and so is \(p(C_2)\), but since \(x\sim y\) we have that \(p(C_1)\cap p(C_2)\neq \varnothing\) so \(X/{\sim}\) consists of a single component, becuase \[p(C_1)\cup p(C_2) = p(C_1\cup C_2)=p(X)=X/{\sim},\] as wanted.

To show the reverse implication, we use the contrapositive of the statement and show: if we for no \(x\in C_1\) or \(y\in C_2\) have that \(x\sim y\), then \(X/{\sim}\) is not connected. Assume the hypothesis and note that then \(p(C_1)\) and \(p(C_2)\) are then disjoint connected subspaces whose union equal all of \(X/{\sim}\) (since \(p\) is surjective). But then the images of \(C_1\) and \(C_2\) under \(p\) are two components of \(X/{\sim}\), showing that \(X/{\sim}\) is not connected. As wanted.

Cauchy’s integral formula for derivatives

It’s soon exam time, so I’m practicing proofs in complex analysis. Right now that means Cauchy’s integral formula for \(n\)’th derivatives.

Let \(G\) be a domain of the complex numbers and \(f: G\to \mathbb{C}\) a holomorphic function. We first want to show that \(f\) can be expressed as a power series, such that $$f(z)=\sum^\infty_{n=0} a_n(z-a).$$ for some \(a\in\mathbb{C}\), let \(B_\rho(a)\) the largest open ball at \(a\) contained in \(G\). We claim that $$a_n = \frac{1}{2\pi i} \oint\frac{f(z)}{(z-a)^{n+1}}dz.$$ By the Cauchy integral formula we have that, for a fixed \(z_0\in B_\rho(a)\) we have $$f(z_0)=\frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0}$$ and by elementary calculations we can, for \(z\in \partial B_r(a)\), write $$\frac{1}{z-z_0} = \frac{1}{z-a} \frac{1}{1-\frac{z_0 -a}{z-a}}=\frac{1}{z-a}\sum^\infty_{n=0} \left(\frac{z_0-a}{z-a}\right)^n,$$ and from above we then have $$\begin{split} f(z_0)& = \frac{1}{2\pi i} \oint \frac{f(z)}{z-z_0} \\ &=\frac{1}{2\pi i}\oint\sum^\infty_{n=0} \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\ &=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)(z_0-a)^n}{(z-a)^{n+1}}dz\\&=\frac{1}{2\pi i}\sum^\infty_{n=0}\oint \frac{f(z)}{(z-a)^{n+1}}dz(z_0-a)^n\\ &=\sum^\infty_{n=0}a_n (z_0-a)^n,\end{split}$$ as wanted. We see that \(f\) is a power series and thus infinitely complex differentiable, and the derivatives are $$f^{(n)}(a)=\frac{n!}{2\pi i}\oint\frac{f(z)}{(z-a)^{n+1}}dz,$$ as desired.

The observant reader will have noticed that I didn’t check that the sums were uniformly convergent, which is needed in other to switch the sum and integral signs, but this is an easy application of the Weierstraß \(M\)-test.


Berg, C. (2016). Complex analysis. Copenhagen: Department of Mathematical Sciences, University of Copenhagen.