11.1. CONTOUR INTEGRALS 251

Definition 11.1.5 Let γ∗ be an oriented C1 curve having a C1 parametrization γ :[a,b]→ γ∗, then

∫γ∗

f (z)dz ≡∫ b

af (γ (t))γ

′ (t)dt ≡∫ b

aRe(

f (γ (t))γ′ (t))

dt

+i∫ b

aIm(

f (γ (t))γ′ (t))

dt

This is well defined because all the functions are continuous. Then∫

γ∗ f (z)dz is called acontour integral.

Proposition 11.1.6 The above contour integral is well defined and for γ∗ an orientedcurve, f →

∫γ∗ f (z)dz is a complex linear map meaning that for a,b ∈ C,

∫γ∗(a f (z)+bg(z))dz = a

∫γ∗

f (z)dz+b∫

γ∗g(z)dz

Also∫

γ∗ f (z)dz = −∫−γ∗ f (z)dz. In addition to this, if M ≥ max{| f (z)| : z ∈ γ∗} , one

obtains the estimate∣∣∣∫γ∗ f (z)dz

∣∣∣ ≤ML where L is the length of γ∗ defined as∫ b

a |γ ′ (t)|dtfor γ a parametrization for γ∗. This number L is well defined. If fn converges uniformly tof on γ∗, then limn→∞

∫γ

fn (z)dz =∫

γf (z)dz

Proof: The claim about the contour integral being linear is a routine computation fromdoing arithmetic for complex numbers and the above definition. This is obvious for a,breal. In case b = 0 and a = i,

i∫

γ∗f (z)dz ≡ i

(∫ b

aRe(

f (γ (t))γ′ (t))

dt + i∫ b

aIm(

f (γ (t))γ′ (t))

dt)

= i∫ b

aRe(

f (γ (t))γ′ (t))

dt−∫ b

aIm(

f (γ (t))γ′ (t))

dt∫γ∗

i f (z)dz ≡∫ b

a

(iRe

(f (γ (t))γ

′ (t))− Im

(f (γ (t))γ

′ (t)))

dt

which is the same thing because it holds for Riemann sums approximating the variousintegrals.

From consideration of real and imaginary parts, the usual change of variables for-mula holds. If γ, γ̂ are two equivalent parametrizations giving the same orientation, γ :[a,b] → γ∗ and γ̂ : [c,d] → γ∗. I need to show these give the same thing for the con-

tour integral. Let s = γ̂−1 ◦ γ (t) so ds =

(γ̂−1 ◦ γ

)′(t)dt. Also γ (t) = γ̂

(γ̂−1 ◦ γ (t)

)so γ ′ (t) = γ̂

′(

γ̂−1 ◦ γ (t)

)(γ̂−1 ◦ γ

)′(t)

∫ d

cf (γ̂ (s)) γ̂

′ (s)ds =∫ b

af(

γ̂

(γ̂−1 ◦ γ (t)

))γ̂′(

γ̂−1 ◦ γ (t)

)(γ̂−1 ◦ γ

)′(t)dt

=∫ b

af (γ (t))γ

′ (t)dt