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If a holomorphic function $f$ on the unit disc satisfies $|f|\le M$ and $f(a)=0$, the Schwarz-Pick lemma applied to $f/M$ gives $|f(0)|/M\le |a|$. Equality for a nonzero central value forces a scaled automorphism of the unit disk. If $f$ extends past the unit circle and $|f(0)|$ is smaller than its minimum boundary modulus, the maximum modulus principle applied to $1/f$ proves that a zero occurs inside the circle.

The zero set of a function $f$ is the preimage $f^{-1}(\{0\})$, consisting of all its zeros of a function. For a real- or complex-valued continuous function, it is a closed set, because the preimage of the closed singleton $\{0\}$ under a continuous function is closed. A nonzero holomorphic function on a connected open set has isolated zeros by the identity theorem for holomorphic functions.

A nonzero holomorphic function on the unit disc can have exactly a prescribed sequence of zeros, counted with multiplicity, if and only if that sequence is locally finite in the disc. Necessity is the identity theorem for holomorphic functions. For sufficiency, remove a finite number $m$ of zeros at zero and enumerate the others as $a_n$, so $|a_n|\to 1$. Use the Weierstrass elementary factors $E_n(t)=(1-t)\exp ({\sum }_{j=1}^n{t}^j/j)$. On every compact subdisc, eventually $|z/a_n|\le q<1$ and $|\log E_n(z/a_n)|\le q^{n+1}/((n+1)(1-q))$. Hence $z^m{\prod }_n{E}_n(z/a_n)$ converges locally uniformly and has exactly the prescribed zeros. Finite sequences simply give finite products.