58.1. PERIODIC FUNCTIONS 1821

Proof: You can assume ∂Pa contains no poles or zeros of f because if it did, then youcould consider a slightly shifted period parallelogram, Pa′ which contains no new zeros andpoles but which has all the old ones but no poles or zeros on its boundary. By Theorem53.1.3 on Page 1678

12πi

∫∂Pa

zf ′ (z)f (z)

dz =m

∑k=1

zk−m

∑k=1

pk. (58.1.2)

Denoting by γ (z,w) the straight oriented line segment from z to w,∫∂Pa

zf ′ (z)f (z)

dz

=∫

γ(a,a+w1)z

f ′ (z)f (z)

dz+∫

γ(a+w1+w2,a+w2)z

f ′ (z)f (z)

dz

+∫

γ(a+w1,a+w2+w1)z

f ′ (z)f (z)

dz+∫

γ(a+w2,a)z

f ′ (z)f (z)

dz

=∫

γ(a,a+w1)(z− (z+w2))

f ′ (z)f (z)

dz

+∫

γ(a,a+w2)(z− (z+w1))

f ′ (z)f (z)

dz

Now near these line segments f ′(z)f (z) is analytic and so there exists a primitive, gwi (z) on

γ (a,a+wi) by Corollary 51.7.5 on Page 1635 which satisfies egwi (z) = f (z). Therefore,

=−w2 (gw1 (a+w1)−gw1 (a))−w1 (gw2 (a+w2)−gw2 (a)) .

Now by periodicity of f it follows f (a+w1) = f (a) = f (a+w2) . Hence

gwi (a+w1)−gwi (a) = 2mπi

for some integer, m because

egwi (a+wi)− egwi (a) = f (a+wi)− f (a) = 0.

Therefore, from 58.1.2, there exist integers, k, l such that

12πi

∫∂Pa

zf ′ (z)f (z)

dz

=1

2πi[−w2 (gw1 (a+w1)−gw1 (a))−w1 (gw2 (a+w2)−gw2 (a))]

=1

2πi[−w2 (2kπi)−w1 (2lπi)]

= −w2k−w1l ∈M.

From 58.1.2 it followsm

∑k=1

zk−m

∑k=1

pk ∈M.