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When is a power series equal to zero?

Example: Take $\sum_{n=0}^\infty a_n(z-z_0)^n$.

Is this power series equal to zero only at $z=z_0$ if we assume that we have infinitely many nonzero $a_n$?

EDIT: Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically zero?

Proof. Suppose that $z_0\in \Omega$ is a limit point for the sequence $\left\{w_k\right\}_{k=1}^\infty$. and that $f(w_k)=0$. First, we show that f is identically zero in a small disc containing $z_0$. For that, we choose a disc D centered at $z_0$ and contained in $\Omega$, and consider the power series expansion of $f$ in that disc

$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$$

If f is not identically zero, there exists a smallest integer m such that $a_m\neq 0$. But then we can write

$$f(z)=a_m(z-z_0)^m(1+g(z-z_0))$$

where $g(z-z_0)$ converges to 0 as $z\rightarrow z_0$. Taking $z=w_k\neq z_0$ for a sequence of points converging to $z_0$, we get a contradiction since $a_m(w_k-z_0)^m\neq 0$ and $1+g(w_k-z_0)\neq 0$, but $f(x_k)=0$.

Question: What I don't understand is why $1+g(w_k-z_0)\neq 0$?

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    $\begingroup$ No. Consider the power series with $z_0 = 0$, $a_0 = -1$, $a_2 = 1$ and $a_i = 0$ for $i \neq 0, 2$. $\endgroup$ – Michael Albanese Sep 23 '14 at 0:40
  • $\begingroup$ I should reword my question. Say you have infinitely many nonzero $a_n$. $\endgroup$ – Enigma Sep 23 '14 at 0:47
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    $\begingroup$ Because $g(z - z_0)$ converges to $0$ as $z \to z_0$. $\endgroup$ – user14972 Sep 23 '14 at 1:06
  • $\begingroup$ Alright, I am seriously sleep-deprived. I failed to notice that part for the past 15 minutes. Thanks Hurkyl. $\endgroup$ – Enigma Sep 23 '14 at 1:07
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No. Consider $\cos z = \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}z^{2n}$.

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