A nonzero holomorphic hypersurface in complex dimension at least two has no isolated points

The germ of $f$ at $a$ is nonzero: otherwise $f$ would vanish on a neighbourhood of $a$, and [L1] would force $f\equiv 0$ on the domain $U$, contrary to the hypothesis. After translating $a$ to $0$ and applying the invertible complex-linear coordinate change from [L2], which preserves local zero sets and isolatedness, we may therefore assume that $a=0$ and that $f$ is regular in $z_m$ of some order $d$. Because $f(0)=0$, that order satisfies $d\ge 1$.

Step 1.1 and [L2] give a neighbourhood $V\subseteq {\mathbb{C}}^{m-1}$ of $0$ and a radius $r>0$ such that every slice over $z^{\mathrm{\prime }}\in V$ has exactly $d$ zeros in $\mathrm{\mid }\zeta \mathrm{\mid }<r$. Since $m\ge 2$, the parameter space ${\mathbb{C}}^{m-1}$ is nontrivial, so choose $z^{\mathrm{\prime }}\in V\setminus \{0\}$ arbitrarily small. Then there exists $z_m$ with $\mathrm{\mid }{z}_m\mathrm{\mid }<r$ and $f(z^{\mathrm{\prime }},z_m)=0$. Because $z^{\mathrm{\prime }}\mathrm{\ne }0$, this zero is different from the origin.

By taking $z^{\mathrm{\prime }}$ arbitrarily close to $0$ in step 2.1, we obtain zeros of $f$ distinct from $0=a$ arbitrarily close to the origin in the chosen coordinates. Undoing the coordinate change shows that the original point $a$ is a limit point of $Z(f)\setminus \{a\}$.