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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?

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  • $\begingroup$ You need that path to be non constant. $\endgroup$ – zhw. May 12 '16 at 22:50
  • $\begingroup$ Oh yes! A non constant path ! $\endgroup$ – mich95 May 12 '16 at 22:51
  • $\begingroup$ Hmmm ... If the path is non constant, then its range is uncountable. $\endgroup$ – zhw. May 12 '16 at 22:56
  • $\begingroup$ note also that a (non-zero) holomorphic/analytic function on an open $U$ only has isolated zeros on $U$ $\endgroup$ – reuns May 12 '16 at 23:06
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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.

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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 $f^{(0)} = f$).

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  • $\begingroup$ $\gamma$ might not be compact... I think the definition of path means continuous image of a closed interval.. my bad! But, if $\omega \in \gamma$ and you let $z \in gamma$, $(f(z)-f(\omega))/(z-\omega)$ will be zero for $z$ approaching $\omega$ along $\gamma$ and hence $f'(\omega)=0$, and we argue inductively. $\endgroup$ – mich95 May 13 '16 at 0:46
  • $\begingroup$ @mich95 I don't understand whether you agree or disagree with the image being compact $\endgroup$ – user258700 May 13 '16 at 11:47

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