Proving an analytic function is the null function If we assume that $D \subseteq \mathbb{C}$ is a connected open set, and $f: D \to \mathbb{C}$ is analytic , and there exists a continuous path $\gamma$ in $D$ where $f(z)$ vanishes. I suspect that $f(z)$ will vanish at all $z \in D$. I think that for any $\omega \in \gamma$, $f^{(n)}(\omega)=0$. This implies that the power series of $f(z)$ around some $w_{0} \in \gamma$ is the zero power series in some disk of convergence, and hence $f(z)$ vanishes in an open set of $D$, and since $D$ is a connected open set, $f(z)$ is identically $0$. Is this reasonnable?
 A: Your argument is more or less the standard one. If $f(z_i)=0$ for $z_i\in S$, and $S$ has a cluster point $z'$, then $f^{(n)}(z')=0$ since the derivatives can be expressed as limiting along all sequences going to $z'$, and we take take the sequence in $S$, over which it is constant. For example $f'(z')=\lim_{z\to z'}\frac{f(z)-f(z')}{z-z'}=\lim_{z_i\to z'}\frac{f(z_i)-f(z')}{z-z'}=\lim_{z_i\to z'}\frac{0}{z-z'}=0$. This implies the conclusion as you stated.
For an alternate proof, consider the integral $\int_{B_{\epsilon}(y)}\frac{f(z)}{z-y}dz=f(y)$ for $y\in \gamma^{int}$. We have by path independence of the integral, $f(y)=\int_{B_{\epsilon}(y)}\frac{f(z)}{z-y}dz=\int_{\gamma}\frac{f(z)}{z-y}dz=0$. This implies that $f$ vanishes in a open neighborhood, and thus vanishes.
A: Since the image of $\gamma$ is infinite and compact, it has an accumulation point. By the identity theorem $f \equiv 0$ on $D$.
Note that it isn't true that if $w$ is a zero of $f$, then $f^{(n)}(w) = 0$ for all $n \in \Bbb N$ (as you seem to suggest). In fact, it is the contrary that holds (with an exception): if $f(w) = 0$ for some $w \in$ a domain, then there must exist some $n \in \Bbb N$ such that $f^{(n)}(w) \neq 0$ unless $f$ is identically zero on the domain; this is because:
$$\bigcap_{n=0}^{\infty} Z(f^{(n)})$$
is a clopen in the domain (where $Z(h)$ denotes the set of zeros of $h$, and $f^{(0)} = f$).
