I am having problems understanding identity theorem (wikipedia) in complex analysis.
I have a holomorphic function $f(z) = \sin{\frac{\pi}{z-1}}$ defined on the unit disc except for the $1$. Roots of this functions are $$f(z) = \sin{\frac{\pi}{z-1}} = 0 \\\ \frac{\pi}{z-1} = k \pi, k \in \mathbb{Z} \\\ k \neq 0: z-1 = \frac{1}{k} \\\ z = 1+ \frac{1}{k}$$ For $k = -1, -2, -3, -4, ...$ we have a sequence of roots that converges to $1$. So there are infinite many roots in a unit disc.
Doesn't this, according to the identity theorem mean that $f$ should be equal to a zero function?
Well, since the unit disc is not a closed set, the limit of the roots is not in the disc... But what if we take a disc twice the size? So, with radius 2?


1 Answer 1


Even with the disc of radius $2$, $f$ is not holomorphic on that disc because of the essential singularity at $z=1$, so the identity theorem does not apply.

  • $\begingroup$ But it is holomorphic on the disc without the $1$, which is still a connected and open subset, so it is a domain. Therefore I should be able to apply the theorem. $\endgroup$
    – Coupeau
    Sep 9, 2019 at 11:01
  • 1
    $\begingroup$ @Coupeau No you can't because as you noted, the disc without one no longer contains the accumulation point, which is critical for the theorem to work. $\endgroup$ Sep 9, 2019 at 11:02

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