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 6.4 The Fundamental Theorems of Integration


Let f be analytic in the simply connected domain D . The theorems in this section show that an antiderivative F can be constructed by contour integration. A consequence will be the fact that in a simply connected domain, the integral of an analytic function f along any contour joining z[0] to z[1] is the same, and its value is given by F(z[1])-F(z[0]) . Because of this, we will be able to use the antiderivative formulas from Calculus to compute the value of definite integrals.

 

Theorem 6.8 (Indefinite Integrals or Anti-derivatives)

Let f(z) be analytic in the simply connected domain D . If z[0] is a fixed value in D and if C is any contour in D with initial point z[0] and terminal point z, then the function given by

F(z) = int(f(z),z = z[0] .. z) , is analytic in D and `F '(z)` = f(z) .

 

If we set z = z[1] in Theorem 6.8, then we obtain the following familiar result for evaluating a definite integral of an analytic function.

 

Theorem 6.9 (Definite Integrals)

 

Let f(z) be analytic in a simply connected domain D . If z[0] and z[1] are two points in D , then

int(f(z),z = z[0] .. z[1]) = F(z[1])-F(z[0]) , where F(z) is any antiderivative of f(z) .

 

Example 6.17, Page 240. Evaluate the definite integral int(1/(2*sqrt(z)),z = 4 .. 8+6*i) = 1+i .

> f:='f': F:='F': g:='g': z:='z':
f := z -> 1/(2*z^(1/2)):
a := 4:
b := 8+6*I:
g := z -> subs(Z=z,int(f(Z),Z)):
`f(z) ` = f(z);
`a ` = a, ` b ` = b;
`g(z) = `, Int(f(z), z) = g(z);
`g(b) ` = expand(g(b)),` and `,`g(a) ` = expand(g(a));
`g(b) - g(a) ` = g(b) - g(a);
`g(b) - g(a) ` = evalc(g(b) - g(a));
`g(b) - g(a) ` = simplify(g(b) - g(a),power);
Int(f(z), z=a..b) = int(f(z), z=a..b);

`f(z) ` = 1/2*1/(sqrt(z))

`a ` = 4, ` b ` = 8+6*I

`g(z) = `, Int(1/2*1/(sqrt(z)),z) = sqrt(z)

`g(b) ` = sqrt(8+6*I), ` and `, `g(a) ` = sqrt(4)...

`g(b) - g(a) ` = sqrt(8+6*I)-sqrt(4)

`g(b) - g(a) ` = 3+I-sqrt(4)

`g(b) - g(a) ` = 1+I

Int(1/2*1/(sqrt(z)),z = 4 .. 8+6*I) = 1+I

 


 Example 6.18, Page 241. Evaluate the definite integral int(cos(z),z = 1 .. i) = -sin(1)+i*sinh(1) .

> f:='f': F:='F': g:='g': z:='z':
f := z -> cos(z):
a := 1:
b := I:
g := z -> subs(Z=z,int(f(Z),Z)):
`f(z) ` = f(z);
`a ` = a, ` b ` = b;
`g(z) = `, Int(f(z), z) = g(z);
`g(b) ` = expand(g(b)),` and `,`g(a) ` = expand(g(a));
`g(b) - g(a) ` = g(b) - g(a);
`g(b) - g(a) ` = evalc(g(b) - g(a));
Int(f(z), z=a..b) = int(f(z), z=a..b);

`f(z) ` = cos(z)

`a ` = 1, ` b ` = I

`g(z) = `, Int(cos(z),z) = sin(z)

`g(b) ` = I*sinh(1), ` and `, `g(a) ` = sin(1)

`g(b) - g(a) ` = sin(I)-sin(1)

`g(b) - g(a) ` = -sin(1)+I*sinh(1)

Int(cos(z),z = 1 .. I) = -sin(1)+I*sinh(1)

 

Example 6.19 and 6.20, Page 241. Evaluate the definite integral int(1/z,z = C .. ` `) = 2*pi*i ,
where C is the unit circle C : abs(z) = 1 , taken with positive orientation

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

`f(z) ` = 1/z

`C: z(t) ` = exp(I*t)

`f(z(t)) ` = 1/exp(I*t)

`dz = z '(t) dt ` = I*exp(I*t), dt

Int(1/z,z = C .. ` `) = Int(I,t = 0 .. 2*pi)

`The anti-derivative is:`

`g(z) = ` = I*t

`g(2*Pi) ` = 2*I*Pi, ` and `, `g(0) ` = 0

`g(2*Pi) - g(0) ` = 2*I*Pi

Int(1/z,z = C .. ` `) = 2*I*Pi

 

End of Section 6.4.