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.
