4.2: Complex Integration
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The magic and power of calculus ultimately rests on the amazing fact that differentiation and integration are mutually inverse operations. And, just as complex functions enjoy remarkable differentiability properties not shared by their real counterparts, so the sublime beauty of complex integration goes far beyond its real progenitor.
Peter J. Oliver
Contour integral
Consider a contour C parametrized by z(t)=x(t)+iy(t) for a≤t≤b. We define the integral of the complex function along C to be the complex number
∫Cf(z)dz=∫baf(z(t))z′(t)dt. (1)
Here we assume that f(z(t)) is piecewise continuous on the interval a≤t≤b and refer to the function f(z) as being piecewise continuous on C. Since C is a contour, z′(t) is also piecewise continuous on a≤t≤b; and so the existence of integral (1) is ensured.
The right hand side of (1) is an ordinary real integral of a complex-valued function; that is, if w(t)=u(t)+iv(t), then
∫baw(t)dt=∫bau(t)dt+i∫bav(t)dt (2)
Now let us write the integrand
f(z)=u(x,y)+iv(x,y)
in terms of its real and imaginary parts, as well as the differential
dz=dzdtdt=(dxdt+idydt)dt=dx+idy
Then the complex integral (1) splits up into a pair of real line integrals:
∫Cf(z)dz=∫C(u+iv)(dx+idy)=∫C(udx−vdy)+i∫C(vdx+udy). (3)
Example 4.2.1
Let’s evaluate ∫Cf(ˉz)dz, where C is given by x=3t, y=t2, y=t2, with −1≤t≤4.

Here we have that C is z(t)=3t+it2. Therefore, with the identification f(z)=ˉz we have
f(z(t))=¯3t+it2=3t−it2.
Also, z′(t)=3+2it, and so the integral is
∫Cˉzdz=∫4−1(3t−it2)(3+2it)dt=∫4−1(2t3+9t+3t2i)dt=∫4−1(2t3+9t)dt+i∫4−13t2dt=(12t4+92t2)|4−1+it3|4−1=195+65i.
Example 4.2.2
Now let’s evaluate ∫C1zdz, where C is the circle x=cost, y=sint, with 0≤t≤2π.

In this case C is z(t)=cost+isint=eit,
f(z(t))=1eit
and, z′(t)=ieit. Thus
∫C1zdz=∫2π0(e−it)ieitdt=i∫2π0dt=2πi.
Numerical evaluation of complex integrals
Exploration 1
Use the following applet to explore numerically the integral
∫Cˉzdz
with different contours C:
- Line segments.
- Semicircles.
- Circles, positively and negatively oriented.
You can also change the domain coloring plotting option. Drag the points around and observe carefully what happens. Then solve Exercise 1 below.
INTERACTIVE GRAPH
The arrows on the contours indicate direction.
Exercise 4.2.1
Exercise 1: Use definition (1) to evaluate ∫Cˉzdz, for the following contours C from z0=−2i to z1=2i:
- Line segment. That is, z(t)=−2i(1−t)+2it, with 0≤t≤1.
- Right-hand semicircle. That is, z(θ)=2eiθ with −π2≤θ≤π2.
- Left-hand semicircle. That is, z(θ)=−2e−iθ with 0≤θ≤π.
Use the applet to confirm your results.
What conclusions (if any) can you draw about the function ˉz from this?
Exploration 2
Now use the applet below to explore numerically the integrals
∫C(z2+z)dz; ∫C1z2dz
with different contours C (line segments, semicircles, and circles). Drag the points around and observe carefully what happens. You can select the functions z^2+z
or 1/z^2
from the list at the left-top corner. Then solve Exercises 2 and 3.
Exercise 4.2.2
Exercise 2: Consider the integral
I1=∫C(z2+z)dz.
Use the applet to analize the value of I1 in the following cases:
- C is any contour from z0=−1−i to z1=1+i.
- C is the circle with center z0 and radius r>0, |z−z0|=r; positively or negatively oriented. In this cases select
Circle ↺
orCircle ↻
.
What conclusions (if any) can you draw about the value of I1 and the function z2+z from this?
Exercise 4.2.3
Exercise 3: Now considering integral
I2=∫C1z2dz.
First, in the applet select the function f(z)=1/z^2
. Then analize the values of I2 in the following cases:
- C is any contour from z0=−i to z1=i. What happens when you select
Line Segment
in the applet? What happens when you selectSemicircles
? - C is the circle with center z0 and radius r>0, |z−z0|=r; positively or negatively oriented. In this case select
Circle ↺
orCircle ↻
. What happens if z=0 is inside or outside the circle? What happens if z=0 lies on the contour, e.g. when \(z0=1\) and r=1?
What conclusions (if any) can you draw about the value of I2 and the function 1z2 from this?
Antiderivatives
Although the value of a contour integral of a function f(z) from a fixed point z0 to a fixed point z1 depends, in general, on the path that is taken, there are certain functions whose integrals from z0 to z1 have values that are independent of path, as you have seen in Exercises 2 and 3. These examples also illustrate the fact that the values of integrals around closed paths are sometimes, but not always, zero. The next theorem is useful in determining when integration is independent of path and, moreover, when an integral around a closed path has value zero. This is known as the complex version of the Fundamental Theorem of Calculus.
Theorem 4.2.1
Let f(z)=F′(z) be the derivative of a single-valued complex function F(z) defined on a domain Ω⊂C. Let C be any countour lying entirely in Ω with initial point z0 and final point z1. Then
∫Cf(z)dz=F(z)|z1z0=F(z1)−F(z0).
Proof: This follows from definition (1) and the chain rule. That is
∫Cf(z)dz=∫CF′(z(t))dzdtdt=∫baddtF(z(t))dt=F(z(b))−F(z(a))=F(z1)−F(z0)
where z0=za and z1=zb are the endpoints of the contour C. ■