53.1. ROUCHE’S THEOREM AND THE ARGUMENT PRINCIPLE 1679

number of zeros of f counted according to the order of the zero with a similar definitionholding for Pf . Thus the conclusion of the argument principle is.

12πi

∫γ

f ′ (z)f (z)

dz = Z f −Pf

Rouche’s theorem allows the comparison of Zh−Ph for h = f ,g. It is a wonderful andamazing result.

Theorem 53.1.4 (Rouche’s theorem)Let f ,g be meromorphic in an open set Ω. Also sup-pose γ∗ is a closed bounded variation curve with the property that for all z /∈Ω,n(γ,z) = 0,no zeros or poles are on γ∗, and for all z ∈ Ω,n(γ,z) either equals 0 or 1. Let Z f and Pfdenote respectively the numbers of zeros and poles of f , which have the property that thewinding number equals 1, counted according to order, with Zg and Pg being defined simi-larly. Also suppose that for z ∈ γ∗

| f (z)+g(z)|< | f (z)|+ |g(z)| . (53.1.4)

ThenZ f −Pf = Zg−Pg.

Proof: From the hypotheses,∣∣∣∣1+ f (z)g(z)

∣∣∣∣< 1+∣∣∣∣ f (z)g(z)

∣∣∣∣which shows that for all z ∈ γ∗,

f (z)g(z)

∈ C\ [0,∞).

Letting l denote a branch of the logarithm defined on C\ [0,∞), it follows that l(

f (z)g(z)

)is a

primitive for the function,( f/g)′

( f/g)=

f ′

f− g′

g.

Therefore, by the argument principle,

0 =1

2πi

∫γ

( f/g)′

( f/g)dz =

12πi

∫γ

(f ′

f− g′

g

)dz

= Z f −Pf − (Zg−Pg) .

This proves the theorem.Often another condition other than 53.1.4 is used.

Corollary 53.1.5 In the situation of Theorem 53.1.4 change 53.1.4 to the condition,

| f (z)−g(z)|< | f (z)|

for z ∈ γ∗. Then the conclusion is the same.

Proof: The new condition implies∣∣∣1− g

f (z)∣∣∣< ∣∣∣ g(z)

f (z)

∣∣∣ on γ∗. Therefore, g(z)f (z) /∈ (−∞,0]

and so you can do the same argument with a branch of the logarithm.