Theorems of Morera and Liouville and Extensions

Module

for

6.6 The Theorems of Morera and Liouville and Extensions

In this section we investigate some of the qualitative properties of analytic and harmonic functions. Our first result shows that the existence of an antiderivative for a continuous function is equivalent to the statement that the integral of f(z) is independent of the path of integration. This result stated in a form that will serve as a converse to the Cauchy-Goursat theorem.

Theorem 6.13 (Morera's Theorem). Let f(z) be a continuous function in a simply connected domain D. If for every closed contour in D, then f(z) is analytic in D.

Proof.

Proof of Theorem 6.13 is in the book.
Complex Analysis for Mathematics and Engineering

Cauchy's integral formula show how the value can be represented by a certain contour integral. If we choose the contour of integration C to be a circle with center , then we can show that the value is the integral average of the values of f(z) at points z on the circle C.

Theorem 6.14 (Gauss's Mean Value Theorem). If f(z) is analytic in a simply connected domain D that contains the circle , then

.

Proof.

Proof of Theorem 6.14 is in the book.
Complex Analysis for Mathematics and Engineering

We now prove an important result concerning the modulus of an analytic function.

Theorem 6.15 (Maximum Modulus Principle). Let f(z) be analytic and nonconstant in the bounded domain D. Then does not attain a maximum value at any point in D.

Proof.

Proof of Theorem 6.15 is in the book.
Complex Analysis for Mathematics and Engineering

We sometimes state the maximum modulus principle in the following form.

Theorem 6.16 (Maximum Modulus Principle). Let f(z) be analytic and nonconstant in the bounded domain D. If f(z) is continuous on the closed region R that consists of D and all of its boundary points B, then assumes its maximum value, and does so only at point(s) on the boundary B.

Proof.

Proof of Theorem 6.16 is in the book.
Complex Analysis for Mathematics and Engineering

Example 6.26.  Let  .  If we set our domain D to be  ,  then f(z) is continuous on the closed region  .  Prove that

            ,

and this value is assumed by f(z) at a point    on the boundary of D.

Solution.  From the triangle inequality and the fact that   in D, it follows that

(6-58)            .

If we choose  ,  where  ,  then

             [Graphics:Images/LiouvilleMoreraGaussMod_gr_22.gif]

so the vectors and lie on the same ray through the origin.  This is the requirement for the Inequality (6-58) to be an equality (see Exercise 19 in Section 1.3).
Hence  ,  and the result is established.

Explore Solution 6.26.

Extra Example 1. Let . If we set our domain D to be , then f(z) is continuous on the closed region . Show that , and this value is assumed by at a point on the boundary of D.

Explore Solution for Extra Example 1.

Extra Example 2. Let . If we set our domain D to be , then f(z) is continuous on the closed region . Show that , and this value is assumed by at a point on the boundary of D.

Explore Solution for Extra Example 2.

Theorem 6.17 (Cauchy's Inequalities). Let f(z) be analytic in the simply connected domain D that contains the circle . If holds for all points , then

for .

Proof.

Proof of Theorem 6.17 is in the book.
Complex Analysis for Mathematics and Engineering

Theorem 6.18 shows that a nonconstant entire function cannot be a bounded function.

Theorem 6.18 (Liouville's Theorem). If f(z) is an entire function and is bounded for all values of z in the complex plane, then f(z) is constant.

Proof.

Proof of Theorem 6.18 is in the book.
Complex Analysis for Mathematics and Engineering

Example 6.27. Show that the function sin(z) is not a bounded function.

Solution. We established this characteristic with a somewhat tedious argument in Section 5.4. All we need do now is observe that f(z) is not constant, and hence it is not bounded.

Explore Solution 6.27.

We can use Liouville's theorem to establish an important theorem of algebra.

Theorem 6.19 (Fundamental Theorem of Algebra). If P(z) is a polynomial of degree , then P(z) has at least one zero.

Proof.

Proof of Theorem 6.19 is in the book.
Complex Analysis for Mathematics and Engineering

Corollary 6.4.  Let P(z) be a polynomial of degree .  Then P(z) can be expressed as the product of linear factors.  That is,



where    are the zeros of P(z) counted according to multiplicity an A is a constant.