COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
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CHAPTER 6 COMPLEX INTEGRATION

Section 6.2 Contours and Contour Integrals

In Section 6.1 we learned how to evaluate integrals of the form , where was complex-valued and was an interval on the real axis (so that was real, with ). In this section we shall define and evaluate integrals of the form , where is complex-valued and is a contour in the plane (so that is complex, with ). Our main result is Theorem 6.1, which will show how to transform the latter type of integral into the kind we investigated in Section 6.1.

We will use concepts first introduced in Section 1.6. Recall that to represent a curve in the plane we use the parametric notation

, for a t b ,

where and are continuous functions. We will place a few more restrictions on the type of curve that we will be studying. The following discussion will lead to the concept of a contour, which is a type of curve that is adequate for the study of integration.

Recall that is simple if it does not cross itself, which means whenever . A curve with the property that is a closed curve . If is the only point of intersection, then we say that is a simple closed curve . As the parameter increases from the value to the value , the point starts at the initial point , moves along the curve , and ends up at the terminal point . If is simple, then moves continuously from to as increases, and the curve is given an orientation , which we indicate by drawing arrows along the curve.

The complex-valued function is said to be differentiable on if both and are differentiable for a t b.

Here we require the one-sided derivatives of and to exist at the endpoints of the interval. As we saw in Section 6.1, the derivative is

for a t b.

The curve is said to be smooth if is continuous and nonzero on the interval. If is a smooth curve, then has a nonzero tangent vector at each point , which is given by the vector .

The opposite curve traces out the same set of points in the plane but in the reverse order, and it has the parametrization

, for t .

A curve that is constructed by joining finitely many smooth curves end to end is called a contour . Let denote smooth curves such that the terminal point of the curve coincides with the initial point of for . We express the contour by the equation

A synonym for contour is path .

Example 6.5, Page 208. Let us find a parameterization of the polygonal path from to
consisting of the three line segments:
Segment 1 from to ,
Segment 2 from to ,
Segment 3 from to .

> t:='t':x0:='x0':x1:='x1':y0:='y0':y1:='y1':z:='z':Z:='Z':Z1:='Z1':
z0 := - 1 + I:
z1 := - 1:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z1 := t -> x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
`Find the segment from -1+i to -1.`;
z[0] = Z1(0),` and `,z[1] = Z1(1);
Z[1](t) = Z1(t);

> t:='t':x0:='x0':x1:='x1':y0:='y0':y1:='y1':z:='z':Z:='Z':Z2:='Z2':
z0 := - 1:
z1 := 1 + I:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z2 := t -> x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
`Find the segment from -1 to 1+i.`;
z[0] = Z2(0),` and `,z[0] = Z2(1);
Z[2](t) = Z2(t);

> t:='t':x0:='x0':x1:='x1':y0:='y0':y1:='y1':z:='z':Z:='Z':Z3:='Z3':
z0 := 1 + I:
z1 := 3 - I:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z3 := t -> x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
`Find the segment from 1+i to 3-i.`;
z[0] = Z3(0),` and `,z[1] = Z3(1);
z[3](t) = Z3(t);

We are now ready to define the integral of a complex function along a contour in the plane with initial point and terminal point . Our approach is to mimic what is done in calculus. We create a partition of points that proceed along from and and form the differences for . Between each pair of partition points and we select a point on , where the function evaluated.

Definition 6.2: Complex integral of f(z)

= ,

provided the limit exists in the sense previously discussed.

Example 6.6, Page 209. Use a Riemann sum to construct an approximation for the contour
integral where is a the line segment joining the point to .

> a:='a':b:='b':C:='C':f:='f':F:='F':n:='n':z:='z':Z:='Z':
F := z -> exp(z):
`f(z) ` = F(z);
a := 0:
b := 2 + I*Pi/4:
n := 8:
`a ` = a, ` b ` = b, ` n ` = n;
dz := (b-a)/8:
Z := k -> a + k*dz:
c := k -> simplify((Z(k-1)+Z(k))/2):
Sum(f(c[k])*Delta*z[k]) = Sum(F(c(k))*dz);

> RS := sum(F(c(k))*dz, k=1..8):
`The Riemann sum is:`;
Sum(f(c[k])*Delta*z[k],k=1..8) = RS;
`The Riemann sum is:`;
Sum(f(c[k])*Delta*z[k],k=1..8) = evalf(RS);

Theorem 6.1 Suppose is a continuous complex-valued function defined on a set containing the contour .

Let be any parameterization of for a t b . Then

.

Example 6.7, Page 212. Evaluate the contour integral
where is a the line segment joining the point to .

> f:='f': F:='F': g:='g': t:='t': T:='T': z:='z': z1:='z1':
f := z -> exp(z):
F := t -> exp((2 + I*Pi/4)*t):
`f(z) ` = f(z);
a := 0:
b := 2 + I*Pi/4:
`a ` = a, ` b ` = b;
z := t -> a + (b - a)*t:
`C: z(t) ` = z(t);
`f(z(t)) = `,F(t) = evalc(F(t));
z1 := t -> subs(T=t,diff(z(T), T)):
Int(f(z),z=C..``) = Int(F(t)*diff(z(t),t),t=0..1);
`The anti-derivative is:`;
g := t -> subs(T=t,int(F(T)*z1(T), T)):
`g(t) ` = g(t);
g1 := g(1):
g0 := g(0):
`g(1) ` = g1, ` g(0) ` = g0;
`g(1) - g(0) ` = g1 - g0;
Int(f(z),z=C..``) = g1 - g0;
Int(f(z),z=C..``) = evalc(g1 - g0);
Int(f(z),z=C..``) = evalf(g1 - g0);

An alternate method is to use the integral of the real and imaginary parts of .
WARNING. Because is a constant for this example, we took it outside the integral!
Also, this computation is very memory intensive !

> f:='f': F:='F': g:='g': t:='t': T:='T': u:='u': v:='v': z:='z':
f := z -> exp(z):
u := t -> exp(2*t)*cos(1/4*t*Pi):
v := t -> exp(2*t)*sin(1/4*t*Pi):
`f(z) ` = f(z);
a := 0:
b := 2 + I*Pi/4:
`a ` = a, ` b ` = b;
z := t -> a + (b - a)*t:
`C: z(t) ` = z(t);
`f(z(t)) = `,f(z(t)) = u(t) + I*v(t);
z1 := diff(z(t), t):
Int(f(z),z=C..``) = Int((u(t)+I*v(t))*diff(z(t),t),t=0..1);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(u(T), T)))*z1 +
I*simplify(subs(T=t,int(v(T), T)))*z1:
`g(t) ` = g(t); ` `;
g1 := simplify(g(1)):
g0 := simplify(g(0)):
`g(1) ` = g1,` and `,
`g(0) ` = g0;
`g(1) - g(0) ` = expand(g1 - g0); ` `;
Int(f(z),z=C..``) = expand(g1 - g0);
Int(f(z),z=C..``) = evalf(g1 - g0);

Example 6.8, Page 212. Evaluate the contour integral dz
where is a the upper semicircle with radius centered at .

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> 1/(z - 2):
`f(z) ` = f(z);
z := t -> 2 + exp(I*t):
`C: z(t) ` = z(t);
`f(z(t)) ` = f(z(t));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*diff(z(t),t),t=0..2*pi);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(f(z(T))*z1(T), T))):
`g(t) ` = g(t);
g1 := g(2*Pi):
g0 := g(0):
`g(2*Pi) - g(0) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

Example 6.9, Page 214. Evaluate the contour integrals of over a line segment and portion of a parabola.
(a) Use the straight line connecting the points and .

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> z:
`f(z) ` = f(z);
z := t -> 2*t + 1 + I*t:
`C: z(t) ` = z(t);
`f(z(t)) ` = f(z(t));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*z1(t),t=-1..1);
Int(f(z),z=C..``) = Int(evalc(f(z(t))*z1(t)),t=-1..1);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(f(z(T))*z1(T), T))):
`g(t) ` = evalc(g(t));
g1 := g(1):
g0 := g(-1):
`g(1) - g(-1) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

(b) Use the portion of a parabola connecting the points and .

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> z:
`f(z) ` = f(z);
z := t -> t^2 + 2*t + I*t:
`C: z(t) ` = z(t);
`f(z(t)) ` = f(z(t));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*z1(t),t=-1..1);
Int(f(z),z=C..``) = Int(evalc(f(z(t))*z1(t)),t=-1..1);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(f(z(T))*z1(T), T))):
`g(t) ` = evalc(g(t));
g1 := g(1):
g0 := g(-1):
`g(1) - g(-1) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

Which is the same as the value for the other path.

Example 6.*, Similar to the above example, but using .
Evaluate the contour integrals of over a line segment and portion of a parabola.
(a) Use the straight line connecting the points and .

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> conjugate(z):
`f(z) ` = f(z);
z := t -> 2*t + 1 + I*t:
`C: z(t) ` = z(t);
`f(z(t)) ` = evalc(f(z(t)));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*z1(t),t=-1..1);
Int(f(z),z=C..``) = Int(evalc(f(z(t))*z1(t)),t=-1..1);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(evalc(f(z(T))*z1(T)), T))):
`g(t) ` = evalc(g(t));
g1 := g(1):
g0 := g(-1):
`g(1) - g(-1) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

(b) Use the portion of a parabola connecting the points and .

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> conjugate(z):
`f(z) ` = f(z);
z := t -> t^2 + 2*t + I*t:
`C: z(t) ` = z(t);
`f(z(t)) ` = evalc(f(z(t)));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*z1(t),t=-1..1);
Int(f(z),z=C..``) = Int(expand(evalc(f(z(t))*z1(t))),t=-1..1);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(evalc(f(z(T))*z1(T)), T))):
`g(t) ` = evalc(g(t));
g1 := g(1):
g0 := g(-1):
`g(1) - g(-1) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

Which is different from the value for the other path.

Example 6.10, Page 215.
(a) Show that , where is the semicircular path from to in the upper half plane.
(b) Show that , where is the polygonal path from to to to .

(a) Use the semicircular path from to in the upper half plane.

> dz :='dz':f:='f':F:='F':g:='g':t:='t':T:='T':z:='z':Z:='Z':z1:='z1':
f := z -> conjugate(z):
`f(z) ` = f(z);
z := t -> - cos(t) + I*sin(t):
`C: z(t) ` = z(t);
`f(z(t)) ` = evalc(f(z(t)));
z1 := t -> subs(T=t,diff(z(T), T)):
`dz = z '(t) dt ` = z1(t), `dt`;
Int(f(z),z=C..``) = Int(f(z(t))*z1(t),t=0..pi);
Int(f(z),z=C..``) = Int(expand(evalc(f(z(t))*z1(t))),t=0..pi);
`The anti-derivative is:`;
g := t -> simplify(subs(T=t,int(evalc(f(z(T))*z1(T)), T))):
`g(t) ` = evalc(g(t));
g1 := g(Pi):
g0 := g(0):
`g(Pi) - g(0) ` = expand(g1 - g0);
Int(f(z),z=C..``) = expand(g1 - g0);

(b) Show that , where is the polygonal path from to to to .

Find a parameterization of the polygonal path from to consisting of the three line segments:
Segment 1 from to ,
Segment 2 from to ,
Segment 3 from to .

> t:='t':x0:='x0':x1:='x1':y0:='y0':y1:='y1':z:='z':
z0:='z0':z1:='z1':Z1:='Z1':Z2:='Z2':Z3:='Z3':
z0 := - 1:
z1 := - 1 + I:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z1 := x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
z0 := - 1 + I:
z1 := 1 + I:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z2 := x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
z0 := 1 + I:
z1 := 1:
x0 := Re(z0): y0 := Im(z0): x1 := Re(z1): y1 := Im(z1):
Z3 := x0 + (x1-x0)*t + I*(y0 + (y1-y0)*t):
C[1], ` is given by `,z[1](t) = Z1;
C[2], ` is given by `, z[2](t) = Z2;
C[3], ` is given by `, z[3](t) = Z3; ` `;

> f:='f': F:='F': g:='g': t:='t': z:='z':
f := z -> conjugate(z):
F(z[1](t)) = evalc(f(Z1)),` and `,
diff(z[1](t),t) = evalc(diff(Z1,t));
w1 := evalc(f(Z1)*diff(Z1,t)):
F(z[1](t))*diff(z[1](t),t) = w1; ` `;
F(z[2](t)) = evalc(f(Z2)),` and `,
diff(z[2](t),t) = evalc(diff(Z2,t));
w2 := evalc(f(Z2)*diff(Z2,t)):
F(z[2](t))*diff(z[2](t),t) = w2; ` `;
F(z[3](t)) = evalc(f(Z3)),` and `,
diff(z[3](t),t) = evalc(diff(Z3,t));
w3 := evalc(f(Z3)*diff(Z3,t)):
F(z[3](t))*diff(z[3](t),t) = w3; ` `;

> s1 := int(w1,t):
Int(F(z[1](t))*diff(z[1](t),t),t) = s1;
s2 := int(w2,t):
Int(F(z[2](t))*diff(z[2](t),t),t) = s2;
s3 := int(w3,t):
Int(F(z[3](t))*diff(z[3](t),t),t) = s3;
s := int(w1,t) + int(w2,t) +int(w3,t):
Int(F(z(t))*diff(z(t),t),t) = s;

> z1:='z1': z2:='z2': z3:='z3':
S1 := Int(w1,t=0..1): s1 := int(w1,t=0..1):
Int(F(z[1](t))*diff(z[1](t),t),t=0..1),` = `, S1 = s1;
S2 := Int(w2,t=0..1): s2 := int(w2,t=0..1):
Int(F(z[2](t))*diff(z[2](t),t),t=0..1),` = `, S2 = s2;
S3 := Int(w3,t=0..1): s3 := int(w3,t=0..1):
Int(F(z[3](t))*diff(z[3](t),t),t=0..1),` = `, S3 = s3;
S := Int(w1+w2+w3,t=0..1):
s := int(w1,t=0..1) + int(w2,t=0..1) + int(w3,t=0..1):
Int(F(z(t))*diff(z(t),t),t=0..1),` = `, S = s;
Int(f(z),z=C..``) = s;

Which is different from the value for the other path.

We now give a few important inequalities relating to complex integrals.

Theorem 6.2 (Integral triangle inequality)

If is a continuous function of the real parameter , then

.

Theorem 6.3 (ML inequality)

If is a continuous function on the contour , then

,

where is the length of the contour and is an upper bound for the modulus on , that is for all .

End of Section 6.2.