If $f(z)$ is analytic in an open connected set $\Omega$ and it has a Taylor series $P(z)$ around $z_0$ in $\Omega$ with positive radius of convergence, do we have $f(z) = P(z)$ for all $z$ in $\Omega$?

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    $\begingroup$ Only if $\Omega$ is contained in the region where $P(z)$ converges. $\endgroup$ – arkeet Oct 1 '16 at 19:15
  • $\begingroup$ Can you give a counterexample? What if we assume that $\Omega$ differs from a simply connected open set by a discrete set? $\endgroup$ – Vik78 Oct 1 '16 at 19:16
  • $\begingroup$ $f(z)=\frac1{z-1}$ is analytic on $\Bbb C\setminus\{1\}$, but the Taylor series around $z_0=0$ does not converge for $|z|>1$. Hence we do not have $f(z)=P(z)$ for all $z\in\Omega$. $\endgroup$ – Hagen von Eitzen Oct 1 '16 at 19:20
  • $\begingroup$ In Hagen's example we can also take $\Omega = \{z : \operatorname{Re}z < 1\}$; then it is simply connected but again $P$ doesn't converge on all of $\Omega$. $\endgroup$ – arkeet Oct 1 '16 at 19:24
  • $\begingroup$ Ah yeah, thanks a lot. $\endgroup$ – Vik78 Oct 1 '16 at 19:25

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