The set of zeros of an holomorphic function is closed and discrete

Let $X$ a Banach space, $f:U\subseteq \mathbb{C}\to X$ an holomorphic function non constant and $N:=\left\{z\in\mathbb{C}:f(z)=0\right\}$. Show that $N$ is closed and discrete.

My idea:

The function $f$ is holomorphic, particulary $f$ is continuos. Then $f^{-1}(\{0\})$ is colsed. Now, suppose that $N$ is not discrete, that is, $N$ has a limit point $\omega\in N$. The function $g=0$ satisfy that $$f(x)=g(x),\hspace{0.2cm}\text{if }x\in N.$$ Finally, $N$ has at least one limit point and cause of the identity principle $$f(x)=g(x)=0\hspace{0.1cm},\text{if }x\in U.$$

Questions:

1. Is it right?
2. Can I use the hypothesis $f$ non zero instead of non constant?

Thanks.

• 1.Yes, that looks right and 2. yes, if $f$ is not the zero function and constant $N=\emptyset$ which is closed and discrete. – Peter Melech Sep 7 '18 at 9:54
• Hint $:$ Since $f$ is non constant the zeros of $f$ has no limit point by identity theorem. – Dbchatto67 Sep 7 '18 at 9:55
• 1. Right , 2. Right too. – Dbchatto67 Sep 7 '18 at 9:57

Identity Principle is for complex valued functions. Here you have to take $x^{*} \in X^{*}$ (where $X^{*}$ is the dual of $X$) and apply Identity Principle for $x^{*} \circ f$ to show that $x^{*} \circ f (z)=0$ for all $z$ for $x^{*}$ (if $N$ has a limit point). This implies $f(z)=0$ for all $z$.
• Could You add to Your answer the definition of $x^{\ast}$ and $B^{\ast}$? – Peter Melech Sep 7 '18 at 10:20
• I suppose You mean $X$ and $X^{\ast}$ instead of $B$...?I upvoted. – Peter Melech Sep 7 '18 at 10:28