Is $f(z) =\sin \frac {\pi} z$ bounded in any nbd of zero and related queries

Question

$f$ is a bounded function defined on $\Bbb{C} - 0$ such that $f(\frac 1n)=0$ for all positive integer $n$. Then (a) $f(0)=0$
(b) $f(z) =0$ for all $z$
(c) $f$ is bounded in any neighbourhood of $0 $
(d) $f(z) $ may assume all complex values except possibly one
Try
Considering the function $f(z) =\sin \frac {\pi} z$ option (a) and (b) is discarded. Now the zeroes of $f(z) =\sin \frac {\pi} z$ are $\frac 1n$ having limit point $0$. So $f(z) =\sin \frac {\pi} z$ has essential singularity at $0$.so it cannot be entire hence option (d) discarded. So I think option (c) true. But I can't prove this option. Please help! Thanks in advance.