Zeros of an analytic function on a domain D.

Question

Suppose $f(z)$ is analytic on a domian $D$ and $z_{0}\in D$. Show that if $f^{(m)}(z_{0})=0$ for $m\geq 1$, then $f(z)$ is constant on $D$.

Maybe this can be proved with the Uniqueness Principle but I need a hint.

Answer 1

There is $r>0$ such that $U=\{z \in \mathbb C: |z-z_0|<r\} \subseteq D$ and

$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ for all $ z \in U$.

Since $a_n=\frac{f^{(n)}(z_0)}{n!}$ for $n \ge 0$, we get $f(z)=a_0$ for all $z \in U$. The identity theorem now gives that $f(z)=a_0$ for all $z \in D$.

Answer 2

Let $h=f-f(z_{0})$, then $h^{(m)}=0$ for all $m=0,1,2,...$, and let $S_{m}=\{w\in D:h^{(m)}(w)=0\}$, then $S_{m}$ is closed in $D$, so is $\displaystyle\bigcap_{m\geq 0} S_{m}$. Try to argue that $\displaystyle\bigcap_{m\geq 0}S_{m}$ is open in $D$. Since $z_{0}\in\displaystyle\bigcap_{m\geq 0}S_{m}$, then $D=\displaystyle\bigcap_{m\geq 0}S_{m}$ by the connectedness of $D$, then it is ready to see that $f=f(z_{0})$ throughout the $D$.

Answer 3

Power series expansion shows that $f$ is a constant in a neighborhood of $z_0$. Now $f'$ is analytic in $D$ and it is zero in that neighborhood. Since zeros of $f'$ have a limit point and $D$ is connected $f'=0$ on $D$. Hence $f$ is a constant in any disk contained in $D$. [ $f(b)-f(a)=\int_{[a,b]} f'(z)dz$]. Now $\{z \in D: f(z)=f(z_0)\}$ is open and closed and hence equals $D$.