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Let $f$ be a complex function of the form $ f\left( z \right) = \sum\limits_{ - \infty }^\infty {a_j z^j } $

If $f(z) = 0$, $ \forall z:\left| z \right| < 1 $, is $f$ the zero function? (in its domain)

I know that functions that has a power series expansion in every point of it´s domain ( let´s suppose also that the domain it´s a open and connected) vanish on neighborhood of point on a domain if and only if , they are the zero function. It´s true that $ f\left( z \right) = \sum\limits_{ - \infty }^\infty {a_j z^j } $ it´s analytic under this definition? enter image description here

This is the defition of having a power series expansion in some point.

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Yes, by analytic continuation, since any Laurent series is analytic where it converges.

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Assume that $f$ is analytic in a domain $\Omega\subset{\mathbb C}$ and that $f$ is identically zero in a neighborhood of $0\in\Omega$. Then $f(z)=0$ for all $z\in\Omega$.

Proof. For any given $z_0\in\Omega$ there is a continuous path $$\gamma:\quad t\mapsto \gamma(t)\in\Omega\qquad(0\leq t\leq1)$$ with $\gamma(0)=0$, $\ \gamma(1)=z_0$.

Assume that $f(z_0)\ne0$, and let $\tau>0$ be the $\inf$ of all $t\geq0$ such that $f\bigl(\gamma(t)\bigr)\ne0$. Then $f$ is not identically $0$ in a neighborhood $U$ of $z_*:=\gamma(\tau)$ and can be written in the form $$f(z)=(z-z_*)^r\ g(z)\qquad (z\in U)\ ,$$ where $r\in{\mathbb N}$, $g$ is analytic in $U$, and $g(z_*)\ne 0$. It follows that $f(z)\ne0$ in a punctured neighborhood of $z_*$, whence $f\bigl(\gamma(t')\bigr)\ne0$ for some $t'<\tau\ $ – a contradiction.

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I have the definition that has locally a power series expansion ( I put the definition in the beginning) I know that this is equivalent that for every point the derivate exist in a neighborhood , but I don´t have that characterization right know – August Feb 25 '12 at 20:02
@August: The above argument works as well using your definition of analyticity. It uses that $\Omega$ is open and connected and that you can take a factor $(z-z_*)^r$ out of the power series of $f$ when $f$ has a zero of order $r$ at $z_*$. – Christian Blatter Feb 26 '12 at 9:16

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