PlanetMath: zeroes of analytic functions are isolated

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zeroes of analytic functions are isolated

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The zeroes of a non-constant analytic function on ${\mathbb{C}}$ are isolated. Let be an analytic function defined in some domain $D \subset {\mathbb{C}}$ and let for some $z_0 \in D$ . Because is analytic, there is a Taylor series expansion for around which converges on an open disk $\vert z-z_0\vert<R$ . Write it as $f(z) = \Sigma_{n=k}^{\infty} a_n (z-z_0)^n$ , with $a_k \ne 0$ and ( is the first non-zero term). One can factor the series so that $f(z) = (z-z_0)^k \Sigma_{n=0}^{\infty} a_{n+k} (z-z_0)^n$ and define $g(z) = \Sigma_{n=0}^{\infty} a_{n+k} (z-z_0)^n$ so that . Observe that is analytic on $\vert z-z_0\vert<R$ .

To show that is an isolated zero of , we must find $\epsilon > 0$ so that is non-zero on $0<\vert z-z_0\vert<\epsilon$ . It is enough to find $\epsilon>0$ so that is non-zero on $\vert z-z_0\vert<\epsilon$ by the relation . Because is analytic, it is continuous at . Notice that $g(z_0)=a_k \ne 0$ , so there exists an $\epsilon > 0$ so that for all with $\vert z-z_0\vert < \epsilon$ it follows that $\vert g(z) - a_k\vert < \frac{\vert a_k\vert}{2}$ . This implies that is non-zero in this set.