Order of a zero of a holomorphic function

If a nonzero holomorphic function has expansion

f (z) = (z - z_{0})^{m} g (z), g (z_{0}) \neq = 0,

then

m

is the order or multiplicity of its zero at

z_{0}

Zero sets in the unit disc

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

E_{n} (t) = (1 - t) exp (\sum_{j = 1}^{n} t^{j} / j)

. On every compact subdisc, eventually

∣ z / a_{n} ∣ \leq q < 1

and

∣ lo g E_{n} (z / a_{n}) ∣ \leq 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.

A simple zero has order one, equivalently

f (z_{0}) = 0

and

f^{'} (z_{0}) \neq = 0

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