# Two fun facts

Well, it seems that my 15 minutes are up, so let’s get back to some complex analysis. Here are two fun facts about holomorphic functions:

Theorem 1 Suppose ${f\in {\mathcal H}(U)}$, and ${U}$ is connected. Then if ${f}$ constant in a neighborhood of some point ${z_0\in U}$. Then ${f}$ is constant on all of ${U}$.

Proof: Since the set ${\{f(z_0)\}}$ is closed, so is it’s preimage. Let ${V}$ denote this preimage. It suffices to show that ${V}$ has no boundary. Then ${V}$ is open, closed, and non-empty. Since ${U}$ is connected, it follows that ${U=V}$.

Suppose ${V}$ has a boundary, and let ${v}$ be some point on the boundary. Expand a power series about ${v}$:

$\displaystyle f(z)=a_0+a_1(z-v)+a_2(z-v)^2+\cdots.$

Since ${f}$ is holomorphic at ${v}$, this power series must have a non-zero radius of convergence so it is valid on some part of the interior of ${V}$. As we know, the only power series with constant image is the constant series ${f(z)=a_0}$. But now this series is valid in a small ball about ${v}$. This contradicts the assumption that ${v}$ was on the boundary of ${V}$, so ${V}$ must be open, proving the theorem.

$\Box$

Theorem 2 Suppose ${f\in{\mathcal H}(U)}$, and ${U}$ is connected. Suppose we have a sequence of points ${z_1,z_2,\dots}$ converging to ${z_\infty\in U}$ such that ${f(z_1)=f(z_2)=\cdots}$. Then ${f}$ is constant on ${U}$

Proof: By continuity of ${f}$, ${f(z_1)=f(z_2)=\cdots=f(z_\infty)}$. Expanding ${f}$ as a power series about ${z_\infty}$, write

$\displaystyle f(z)=a_0+a_1(z-z_\infty)+a_2(z-z_\infty)^2+\cdots$

Since we have a sequence of points converging to ${z_\infty}$ whose ${f}$-values all agree with ${z_\infty}$, we see that the derivative at ${z_\infty}$ must be

$\displaystyle f'(z_\infty)=\lim_{n\rightarrow\infty}\frac{f(z_n)-f(z_\infty)}{z_n-z_\infty}=\lim_{n\rightarrow\infty}\frac{0}{z_n-z_\infty}=0$

Thus, ${f'(z)=0}$ in a neighborhood of ${z_\infty}$, meaning that the power series: ${f'(z)=a_1+2a_2(z-z_\infty)+3a_3(z-z_\infty)^2+\cdots}$ must in fact be constant. Thus, ${a_n=0}$ for ${n>0}$, meaning that ${f(z)=a_0}$ in a neighborhood of ${z_\infty}$.

Now we can apply the previous theorem to conclude that ${f}$ is in fact constant of all of ${U}$.

$\Box$