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whilst following Gamelin's complex analysis book, I came across this statement in the proof for the zeros of $f$ analytic being isolated, eventually used to prove the Uniqueness Principle. I am having a little trouble justifying the statement to myself.

So I am seeking a proof for: If $f$ is analytic and $\exists k \in \mathbb{N}$ such that $f^{(k)}(z_0) \neq 0$, then $\exists \epsilon > 0 \ s.t. f(z) \neq 0$ for all $z \in B(z_0, \epsilon)\setminus\{z_0\}$

This seems to be the key in the proof since its converse would lead to $z_0$ being a limit point of zeros of $f$, which is pretty bad as that contradicts the theorem we're proving.

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  • $\begingroup$ Are you looking to prove this directly from first principles, or with something like Rouch\e's theorem? $\endgroup$
    – anomaly
    Jan 2, 2018 at 23:18
  • $\begingroup$ power series expansion might be a good place to start $\endgroup$ Jan 2, 2018 at 23:18
  • $\begingroup$ @anomaly I'd like to go first principles if possible, I don't think the author went through Rouche's theorem yet in his book at this point $\endgroup$
    – CowNorris
    Jan 2, 2018 at 23:20
  • $\begingroup$ @qbert I was trying that at first also, however I was finding it hard to justify that the remaining infinite polynomial sequence in $(z-z_0)$ must not have further zeros. Could I possibly have a further hint? :P $\endgroup$
    – CowNorris
    Jan 2, 2018 at 23:20
  • $\begingroup$ Can you prove it if $k=0$? $\endgroup$ Jan 2, 2018 at 23:23

2 Answers 2

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Let $k$ be the first $k$ that this happens, then taking the Taylor expansion for $f$ centered at $z_0$ we have

\begin{align*} f(z)&=f^{(k)}(z_0)\frac{(z-z_0)^{k}}{k!}(1-\frac{k!}{f^{(k)}(z_0)}\sum_{i\geq k+1} f^{(i)}(z_0)\frac{(z-z_0)^{i-k}}{i!}) \\ &=f^{(k)}(z_0)\frac{(z-z_0)^{k}}{k!}(1-g(z)) \end{align*} Now, $g(z)$ is continuous and $g(0)=0$. From here the results follows.

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Assume without loss of generality that $z_0 = 0$. Note first that if $f(0) \not = 0$, then the result follows by continuity. Otherwise, we have $f(z) = c_k z^k + c_{k+1} z^{k+1}+\dotsb = z^k(c_k + c_{k+1} z + \dotsb) = z^k g(z)$ on some neighborhood of $0$ for some $k > 0$ with $c_k\not = 0$ and thus $g(0)\not = 0$. Thus $g(z)\not = 0$ on some (smaller) neighborhood of $0$, and we're done.

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