11.6. COUNTING ZEROS, OPEN MAPPING THEOREM 269

1. There exists z ∈U where f (n) (z) = 0 for all n.

2. f is 0 on U.

3. The set of zeros of f has a limit point in U

Proof: 1.⇐⇒ 2. The first two are equivalent by Proposition 11.6.2.2.)⇒ 3.) This is obvious. Since f is 0 everywhere, all derivatives at every point are 0

so every point of B(z0,R) is a limit point of the set of zeros.3.)⇒ 1.)Suppose there exists a limit point z ∈U of the set of zeros. I will show that

f (n) (z) = 0 for all n. By continuity f (z) = 0. Since z is in U , there exists r > 0 such thatB(z,r)⊆U. By Corollary 11.2.11, there are complex numbers ak such that for w∈ B(z,r) ,

f (w) =∞

∑k=1

ak (w− z)k .

If each ak = 0, then at this z, all derivatives of f equal 0. Otherwise,

f (w) = (w− z)m g(w) = (w− z)m∞

∑k=m

ak (w− z)k−m , m > 0

where am ̸= 0. I will show this does not happen. From the above, there exists a sequence ofdistinct zn converging to z where f (zn) = 0. Then 0= f (zn) = (zn− z)m g(zn) so g(zn) = 0.By continuity, g(z) = limn→∞ g(zn) = 0 which requires am = 0 after all. Thus all ak = 0after all, a contradiction. It follows that all three conditions are equivalent.

The counting zeros theorem is as follows:

Theorem 11.6.4 Let f be analytic in an open set containing the closed disk

D(z0,r)≡ {z : |z− z0| ≤ r}

and suppose f has no zeros on the circle C (z0,r) , the boundary of D(z0,r). Then thenumber of zeros of f counted according to multiplicity which are contained in D(z0,r) is

12πi∫

C(z0,r)f ′(z)f (z) dz where C (z0,r) is oriented in the counter clockwise direction.

Proof: There are only finitely many zeros in D(z0,r) . Otherwise, there would exista limit point of the set of zeros z. If z is in B(z0,r) , then by Theorem 11.6.3, f = 0 onD(z0,r). If it is on C (z0,r) , this would contradict having no zeros on the boundary.

Let these zeros be{

z1, ...,zp}

. Consider zk and suppose it is a zero of multiplicity m sof (z) = (z− zk)

m g(z) where g(zk) ̸= 0,m≥ 1. Then

f ′ (z)f (z)

=m(z− zk)

m−1 g(z)+(z− zk)m g′ (z)

(z− zk)m g(z)

=m

(z− zk)+

g′ (z)g(z)

The second term is analytic near zk and so the residue of f ′(z)f (z) is m the number of times

zk is repeated in the list of zeros. Hence doing this for each of the zeros, gives 2πi(p) =∫C(z0,r)

f ′(z)f (z) dz and so p = 1

2πi∫

C(z0,r)f ′(z)f (z) dz.

Obviously the above theorem applies to more general regions than disks, but the maininterest tends to be for balls. Also it generalizes to the situation where there are no poles orzeros on C (z0,r) and finitely many zeros and poles in B(z0,r). In this case, you get a countof the number of zeros minus the number of poles. This more general theorem is calledRouche’s theorem.