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 6.5 Integral Representations for Analytic Functions

We now present some major results in the theory of functions of a complex variable. The first result is known as Cauchy's integral formula and shows that the value of an analytic function can be represented by a certain contour integral. The derivative , , will have a similar representation. In Chapter 7 we will show how the Cauchy integral formulae are used to prove Taylor's theorem, and we will establish the power series representation for analytic functions. The Cauchy integral formulae will also be a convenient tool for evaluating certain contour integrals.

Theorem 6.10 (Cauchy's Integral Formula)

Let be analytic in the simply connected domain , and let be a simple closed positively oriented contour that lies in D.
If is a point that lies interior to , then .

Example 6.21, Page 245. Show that ,
where is the circle : with positive orientation.

> f :='f': F:='F': z:='z': Z:='Z': Zn:='Zn':
w := exp(z)/(z-1):
print(`Find `,Int(w, z=C..``));
Zn := sort([solve(denom(w)=0, z)]):
`Singularity at `, z[0] = Zn[1];
f := Z -> subs(z=Z, w*(z - Zn[1])):
`Use f(z) ` = f(z);

The integral of taken over is:

> `f(z)` = f(z);
`f(z0)` = f(Zn[1]);
val := 2*Pi*I*f(Zn[1]):
Int(w ,z=C..``) = `2 Pi I f(z0)`;
Int(w ,z=C..``) = val;

Example 6.22, Page 245. Show that ,
where is the circle : with positive orientation.

> f :='f': F:='F': z:='z': Z:='Z': Zn:='Zn':
w := sin(z)/(4*z + Pi):
print(`Find `,Int(w ,z=C..``));
Zn := sort([solve(denom(w)=0, z)]):
`Singularity at `, z[0] = Zn[1];
`Singularity at `, z[0] = evalf(Zn[1]);
f := Z -> subs(z=Z, simplify(w*(z - Zn[1]))):
`Use f(z) ` = f(z);

The integral of taken over is:

> `f(z)` = f(z);
`f(z0) ` = f(Zn[1]);
`f(z0) ` = evalc(f(Zn[1]));
val := 2*Pi*I*f(Zn[1]):
Int(w ,z=C..``) = `2 Pi I f(z0)`;
Int(w ,z=C..``) = val;

Example 6.23, Page 245. Show that ,
where is the circle : with positive orientation.
Solution:
We will find that the only singularity that lies inside .

> f :='f': F:='F': z:='z': Z:='Z': Zn:='Zn':
w := exp(I*Pi*z)/(2*z^2 - 5*z + 2):
print(`Find `,Int(w ,z=C..``));
Zn := sort([solve(denom(w)=0, z)]):
`Singularities occur at: ` = Zn;
`The singularity inside C is `,z[0] = Zn[1];
f := Z -> subs(z=Z, simplify(w*(z - Zn[1]))):
`Use f(z) ` = f(z);

We now state a general result that shows how differentiation under the integral sign can be accomplished. The proof can be found in some advanced texts. See, for instance, Rolf Nevanlinna and V. Paatero, Introduction to Complex Analysis , (Reading, Massachusetts: Addison-Wesley Publishing Company, 1969), Section 9.7.

Theorem 6.11 (Leibniz's Rule)

Let be a simply connected domain, and let a t b be an interval of real numbers. Let and its partial derivative , with respect to z, be continuous functions for all z in and all t in I.
Then is analytic for z in D, and .

We now show how we can generalize Theorem 6.10 to give an integral representation for the derivative , . Leibniz's rule is used in the proof, and this method of proof will be a mnemonic device for remembering the upcoming theorem.

Theorem 6.12 (Cauchy's Integral Formulae for Derivatives)

Let be analytic in the simply connected domain , and let be a simple closed positively oriented contour that lies in .
If z is a point that lies interior to C, then .

Example 6.25, Page 247. Show that .
where is the circle : with positive orientation.
Solution:
Since is a singularity of order , multiply the integrand by to get the function .

> f :='f': F:='F': z:='z': Z:='Z': Zn:='Zn':
w := exp(z^2)/(z - I)^4:
print(`Find `,Int(w ,z=C..``));
Zn := sort([solve(denom(w)=0, z)]):
`Singularity of order 4 at `, z[0] = Zn[1];
f := Z -> subs(z=Z, simplify(w*(z - Zn[1])^4)):
`Use f(z) ` = f(z);

The integral of taken over is obtained by applying the
Cauchy integral formula for derivatives.

> d3 := diff(f(z), z$3):
f3 := Z -> subs(z=Z, d3):
`f '''(z) ` = f3(z);
`f '''(z0) ` = f3(Zn[1]);
val := 2*Pi*I/3! * f3(Zn[1]):
Int(w ,z=C..``) = `2 Pi I/3! f '''(z0)`;
Int(w ,z=C..``) = val;

Remark. Sometimes Maple will form the list of values in a different order.

It is always necessary to visually inspect the above results before proceeding.

The integral of taken over is obtained by applying the Cauchy integral formula.

> `f(z0) ` = f(Zn[1]);
`f(z0) ` = evalc(f(Zn[1]));
val := 2*Pi*I*f(Zn[1]):
Int(w ,z=C..``) = `2 Pi I f(z0)`;
Int(w ,z=C..``) = val;

Corollary 6.2 If is analytic in the domain , then all derivatives exists and are analytic in .

Remark. Corollary 6.2 is interesting, as it illustrates a big difference between real and complex functions. It is possible for a real function to have the property that exists everywhere in a domain , but exists nowhere. This corollary states that if a complex function has the property that exists everywhere in a domain , then, remarkably, all derivatives of exist in .

Corollary 6.3 If is a harmonic function at each point in the domain , then all partial derivatives exists and are harmonic functions.

End of Section 6.5.