The question I am trying to answer is:

Suppose $f$ is analytic in the punctured disc $0 < |z| < 1$ except for poles $\{z_n\}$ where: $$\lim_{n \to \infty}z_n = 0$$

Note that $0$ is not an isolated singularity. Show that the range of $f$ in the punctured disc is dense in the complex plane.

I think I have to use a variant of the Casorati-Weierstrass Theorem; possibly Picard's Big Theorem, but I'm not sure how.

Can anyone help?


1 Answer 1


Suppose there is an epsilon ball around $w$ which is disjoint from the image of $f$. Consider $g(z)=\frac{1}{f(z)-w}$ which is analytic except for at $z_n$ and $0$. Since $g$ is bounded above by $\frac{1}{\epsilon}$ these singularities are removable. Hence $g$ extends to an analytic function which is zero at $z_n$ (and at $0$). Hence $g$ is zero on a set with a limit point and so $g$ is identically zero. But this is impossible by definition of $g$.

  • $\begingroup$ So it IS basically an extension of the Casorati-Weierstrass Theorem with more than one singularity? Thank you! $\endgroup$
    – zhn11tau
    Commented Apr 11, 2014 at 23:00

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