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If $u(z)$ is real harmonic and bounded in the punctured disk $0<|z-z_0|<R.$ Show that $\lim_{z\to z_0} u(z)$ exists.

I already know Complex analytic function $f$ which has singularity $z_0$ and bounded on some deleted neighborhood of $z_0$, Then $\lim_{z\to z_0} f(z)$ exists and $f$ can be defined at $z_0$ so that it is analytic at $z_0$.

Seongqjini
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  • if $f(z)$ is holomorphic on the punctured disk and $Re(f(z))$ is bounded, do you get $f(z)$ is holomorphic on the whole disk ? – reuns Jun 17 '16 at 19:53
  • Punctured disk is not simply connected domain. So Analytic function $f$ with $Re(f(z)=u$ does not need to exist. – Seongqjini Jun 18 '16 at 16:04
  • no. if $f(z)$ is analytic on the punctured disk but not at $z=z_0$, then $z_0$ is an isolated singularity, and by the classification theorem : $z_0$ is a pole or an essential singularity. in both cases $Re(f(z))$ is unbounded, so that $Re(f(z))$ bounded $\implies$ $f(z)$ is analytic on the whole disk. $\quad$ note that you can also consider $g(z) = e^{-f(z)}$ which is analytic on the punctured disk and bounded on the whole disk, hence it is analytic on the whole disk (by Riemann's theorem on removable singularities) – reuns Jun 18 '16 at 22:20

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