# Identity theorem and function $f(z) = \sin{\frac{\pi}{z-1}}$ on unit disc

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?

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.
• 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. Sep 9, 2019 at 11:01