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Complex Integration


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 atb. We define the integral of the complex function along C to be the complex number (1)Cf(z)dz=abf(z(t))z(t)dt. Here we assume that f(z(t)) is piecewise continuous on the interval atb 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 atb; 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 (2)abw(t)dt=abu(t)dt+iabv(t)dt

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:

(3)Cf(z)dz=C(u+iv)(dx+idy)=C(udxvdy)+iC(vdx+udy).

Example 1: Let's evaluate Czdz, where C is given by x=3t,y=t2, with 1t4.

Piecewise smooth curve
C: z(t)=3t+it2, with 1t4.

Here we have that C is z(t)=3t+it2. Therefore, with the identification f(z)=z we have f(z(t))=3t+it2=3tit2. Also, z(t)=3+2it, and so the integral is Czdz=14(3tit2)(3+2it)dt=14(2t3+9t+3t2i)dt=14(2t3+9t)dt+i143t2dt=(12t4+92t2)|14+it3|14=195+65i.

Example 2: Now let's evaluate C1zdz, where C is the circle x=cost,y=sint, with 0t2π.

Piecewise smooth curve
C: z(t)=cost+isint=eit, with 0t2π.

In this case C is z(t)=cost+isint=eit, f(z(t))=1eit and, z(t)=ieit. Thus C1zdz=02π(eit)ieitdt=i02πdt=2πi.


Numerical evaluation of complex integrals

Exploration 1

Use the following applet to explore numerically the integral Czdz 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.

The arrows on the contours indicate direction.

Exercise 1: Use definition (1) to evaluate Czdz, for the following contours C from z0=2i to z1=2i:

  1. Line segment. That is, z(t)=2i(1t)+2it, with 0t1.
  2. Right-hand semicircle. That is, z(θ)=2eiθ with π2θπ2.
  3. Left-hand semicircle. That is, z(θ)=2ieiθ with 0θπ.

Use the applet to confirm your results.

What conclusions (if any) can you draw about the function z 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 2: Consider the integral I1=C(z2+z)dz. Use the applet to analyze the value of I1 in the following cases:

  1. C is any contour from z0=1i to z1=1+i.
  2. C is the circle with center z0 and radius r>0, |zz0|=r; positively or negatively oriented. In this cases select Circle ↺ or Circle ↻.

What conclusions (if any) can you draw about the value of I1 and the function z2+z from this?

Exercise 3: Now considering integral I2=C1z2dz. First, in the applet select the function f(z)=1/z^2. Then analyze the values of I2 in the following cases:

  1. C is any contour from z0=i to z1=i. What happens when you select Line Segment in the applet? What happens when you select Semicircles?
  2. C is the circle with center z0 and radius r>0, |zz0|=r; positively or negatively oriented. In this case select Circle ↺ or Circle ↻. 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?


Properties

All the following familiar properties of integrals can be proved directly from the definition given in (2). If w(t)=u(t)+iv(t) and s(t)=υ(t)+iν(t) are complex-valued functions of a real variable t continuous on an interval [a,b], then

(4)abkw(t)dt=kabw(t)dt,k is a complex constant,(5)ab[w(t)+s(t)]dt=abw(t)dt+abs(t)dt,(6)abw(t)dt=acw(t)dt+cbw(t)dtc[a,b],(7)abw(t)dt=baw(t)dt.

From definition (1), and the properties just mentioned above, it also follows immediately that

(8)Cz0f(z)dz=z0Cf(z)dz,z0 is a complex constant,(9)C[f(z)+g(z)]dz=Cf(z)dz+Cg(z)dz

Now consider the contour defined for the integral (1). The contour C consists of the same set of points but with the order reversed so that the new contour extends from z2 to z1, as shown in Figure 3.

Reversed contour
The contour C extends from z2 to z1.

The contour C has a parametric representation z=z(t) with bta. Hence

Cf(z)dz=baf[z(t)]ddtz(t)dt=baf[z(t)]z(t)dt.

If we use the substitution τ=t in the last integral, then

Cf(z)dz=abf[z(τ)]z(τ)dt.

Hence, we obtain the following property

(10)Cf(z)dz=Cf(z)dz.

Consider now a path C, parametrized by z(t) for t[a,b], that consists of a contour C1 from z1 to z2 followed by a contour C2 from z2 to z3, the initial point of C2 being the final point of C1. Then

(11)Cf(z)dz=C1f(z)dz+C2f(z)dz.
Sum of contours
C=C1+C2.

Exercise 4: Use the properties for integrals of functions w(t) to prove (11).

Finally we introduce an inequality involving contour integrals that is extremely important in various applications. Consider C a smooth curve parametrized by z(t)=x(t)+iy(t) with t[a,b]. Suppose that the components x(t) and y(t) of the derivative z(t)=x(t)+iy(t) are continuous on the entire interval [a,b]. Then the real-valued function |z(t)|=[x(t)]2+[y(t)]2 is integrable on [a,b]. According to the definition of arc length from calculus, the length of C is the number L=ab|z(z)|dt.

If C is a contour of length L and f is a piecewise continuous function on C, and M is a nonnegative constant such that |f(z)|M for all points z on C at which f(z) is defined, then (12)|Cf(z)dz|ML

Note that since C is a contour and f is piecewise continuous on C, a number M appearing in the inequality |f(z)|M will always exist. The reason is that the real-valued function |f(z(t))| is continuous on the closed bounded interval [a,b] when f is continuous on C; and this function always reaches a maximum value M on that interval. Therefore |f(z)| has a maximum value on C when f is continuous on it. This is also true when f is piecewise continuous on C.

The property (12) is often used in the theory of complex integration and is sometimes referred to as the ML-inequality.

Example 3: Let C be the contour of the circle |z|=2 from z1=2 to z2=2i that lies in the first quadrant, as shown in Figure 5. We can use inequality (12) to show that |Cdzz21|π3. without evaluating the integral.

ML-inequality example
Circle |z|=2 from z1=2 to z2=2i.

Note that if z is a point of C, then

|z21|||z2|1|=||z|21|=|41|=3

Thus |1z21|=1|z21|13 Also the length of C is 14(4π). Thus, if we take M=13 and L=π, we find that |Cdzz21|ML=π3.

Example 4: We can also use the ML-inequality to find the value of the integral limRCRz1/2z2+1dz where CR is the arc z(t)=Reit, with t[0,π] and z1/2 denotes the branch z1/2=exp(12logz)=reit/2,(r>0,π2<t<3π2) of the square root function.

ML-inequality example
The arc z(t)=Reit, with t[0,π]. The dashed ray indicates the branch cut.

Note that when |z|=R>1, then |z1/2|=|Reit/2|=R and |z2+1|||z2|1|=R21. Thus, at points on CR, we have

|z1/2z2+1|MRwhereMR=RR21.

Since the length of CR is L=πR, using (12), we obtain

|CRz1/2z2+1dz|MRL=πRRR211/R21/R2=πR11R2.

The term on the right tends to zero as R tends to infinity. Hence limRCRz1/2z2+1dz=0.


Antiderivatives

While the value of a contour integral of a function f(z) from a fixed point z0 to a fixed point z1 generally depends on the chosen path, there are certain functions for which the 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 theorem is known as the complex version of the Fundamental Theorem of Calculus and involves the concept of antiderivative (or primitive) of a continuous function f(z) on a domain D, that is, a function F(z) such that F(z)=f(z) for all zD.

Suppose that f(z) is continuous on a domain D. If f(z) has an antiderivative throughout D, then for any contour C lying entirely in D, with initial fixed point z1 and final fixed point z2, we have that (13)Cf(z)dz=F(z)|z1z2=F(z2)F(z1).
This follows from definition (1) and the chain rule. That is Cf(z)dz=CF(z(t))dzdtdt=abddtF(z(t))dt=F(z(b))F(z(a))=F(z2)F(z1) where z1=z(a) and z2=z(b) are the endpoints of the contour C.

As a consequence of the previous Theorem 1, for closed curves we have (14)Cf(z)dz=CF(z)dz=0. for every closed contour C, that is, the endpoints are equal.

If the function f(z) satisfies the hypothesis of Theorem 1, then for all contours C lying in D beginning at z1 and ending at z2 we have expression (13). Hence the result demonstrates that the integral is independent of path. This fact is illustrated in Figure 6.

Path independence
Independent paths forming closed curve

Considering Figure 6, we have C1f(z)dz=C2f(z)dz because

Cf(z)dz=C1f(z)dzC2f(z)dz=0
where the closed contour is C=C1C2. In other words:

If f is continuous and has an antiderivative throughout a domain D, then Cf(z)dz is independent of path.

In addition, we have the following sufficient condition for the existence of an antiderivate:

If f is continuous and Cf(z)dz is independent of path, then f has an antiderivate everywhere in D.
Consider the function (15)F(z)=z0zf(s)ds where s denotes a complex variable, z0 is a fixed points in D, and z represents any point in D. Assuming that f is continuous and Cf(z)dz is independent of path, we wish to prove that F(z)=f(z), that is, (15) is an antiderivate of f in D.

Using properties of the integral we have

(16)F(z+Δz)F(z)=z0z+Δzf(s)dsz0zf(s)ds=zz+Δzf(s)ds.

Since D is a domain, we can choose Δz so that z+Δz is in D. Furthermore, z and z+Δz can be joined by a straight segment, as shown un Figure 7. This is the contour we use in the last integral in (16). For z fixed, we can write

(17)f(z)Δz=f(z)zz+Δzds=zz+Δzf(z)dsorf(z)=1Δzzz+Δzf(z)ds.
Proof
The points z and z+Δz can be joined by a straight segment.

From (16) and (17) we obtain

F(z+Δz)f(z)Δz=1Δzzz+Δz[f(s)f(z)]ds.

Now, since f is continuous, for any ε>0 there exists a δ>0 such that |f(s)f(z)|<ε whenever |sz|<δ. Thus, if we choose z+Δz close enough to z so that |Δz|<δ, it follows from the ML-inequality that

|F(z+Δz)f(z)Δz|=|1Δzzz+Δz[f(s)f(z)]ds|=|1Δz||zz+Δz[f(s)f(z)]ds||1Δz|ε|Δz|=ε.

Therefore, we have proved that

limΔz0F(z+Δz)f(z)Δz=f(z)orF(z)=f(z).

Integrals of Functions with Branch Cuts