# 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 😉

## References

Berg, C. (2016). Complex analysis. 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.

## References

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