= Solution
Let $f$ have the stated <zeros> but not be identically zero. The origin is an <isolated singularity>. If it were a <removable singularity>, the extended <holomorphic function> would have <zeros> accumulating at an interior point. The proof in part (c) would force the extension, and then $f$ on its <domain>, to be identically zero.
If the origin were a <pole> of order $m$, then $g(z)=z^mf(z)$ would have a <holomorphic> extension with $g(0)\ne0$. <Continuity> would make $g$ nonzero on a small disc, so $f$ would have no <zeros> in its punctured disc, again a contradiction. The classification of <isolated singularities> leaves only an \b[essential singularity at the origin]. This argument applies to every function in part (d), not just the particular sine example.
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