Module

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Integral Representations for Analytic Functions

Cauchy Integral Formula

 

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 f(z) can be represented by a certain contour integral.  The  [Graphics:Images/IntegralRepresentationMod_gr_1.gif]  derivative,  [Graphics:Images/IntegralRepresentationMod_gr_2.gif],  will have a similar representation.  In Section 7.2, we use the Cauchy integral formulas to prove Taylor's theorem and also establish the power series representation for analytic functions. The Cauchy integral formulas are a convenient tool for evaluating certain contour integrals.

 

Theorem 6.10 (Cauchy Integral Formula).  Let f(z) be analytic in the simply connected domain D,  and let C be a simple closed positively oriented contour that lies in D.  If  [Graphics:Images/IntegralRepresentationMod_gr_3.gif]  is a point that lies interior to C, then   

            [Graphics:Images/IntegralRepresentationMod_gr_4.gif].   

Proof.

Proof of Theorem 6.10 is in the book.
Complex Analysis for Mathematics and Engineering

 

Example 6.21.  Show that  [Graphics:Images/IntegralRepresentationMod_gr_5.gif],  where C is the circle  [Graphics:Images/IntegralRepresentationMod_gr_6.gif]  with positive orientation.

[Graphics:Images/IntegralRepresentationMod_gr_7.gif]

Solution.  We have  [Graphics:Images/IntegralRepresentationMod_gr_8.gif]  and  [Graphics:Images/IntegralRepresentationMod_gr_9.gif].  The point  [Graphics:Images/IntegralRepresentationMod_gr_10.gif]  lies interior to the circle, so Cauchy's integral formula implies that  

            [Graphics:Images/IntegralRepresentationMod_gr_11.gif],  

and multiplication by  [Graphics:Images/IntegralRepresentationMod_gr_12.gif]  establishes the desired result.

Explore Solution 6.21.

 

Example 6.22.  Show that  [Graphics:Images/IntegralRepresentationMod_gr_27.gif],  where C is the circle  [Graphics:Images/IntegralRepresentationMod_gr_28.gif]  with positive orientation.

[Graphics:Images/IntegralRepresentationMod_gr_29.gif]

Solution.  Here we have  [Graphics:Images/IntegralRepresentationMod_gr_30.gif].  We manipulate the integral and use Cauchy's integral formula to obtain  

            [Graphics:Images/IntegralRepresentationMod_gr_31.gif]  

Explore Solution 6.22.

 

Example 6.23.  Show that  [Graphics:Images/IntegralRepresentationMod_gr_46.gif],  where C is the circle  [Graphics:Images/IntegralRepresentationMod_gr_47.gif]  with positive orientation.

[Graphics:Images/IntegralRepresentationMod_gr_48.gif]

Solution.  We see that  [Graphics:Images/IntegralRepresentationMod_gr_49.gif].  The only zero of this expression that lies in the interior of C is [Graphics:Images/IntegralRepresentationMod_gr_50.gif].  
We set  [Graphics:Images/IntegralRepresentationMod_gr_51.gif]  and use Theorem 6.10 to conclude that  

            [Graphics:Images/IntegralRepresentationMod_gr_52.gif]   

Explore Solution 6.23.

 

Theorem 6.11 (Leibniz's Rule).  Let G be an open set,  and let  [Graphics:Images/IntegralRepresentationMod_gr_68.gif]  be an interval of real numbers.  Let [Graphics:Images/IntegralRepresentationMod_gr_69.gif] and its partial derivative [Graphics:Images/IntegralRepresentationMod_gr_70.gif] with respect to z be continuous functions for all z in G and all t in I.  Then  

            [Graphics:Images/IntegralRepresentationMod_gr_71.gif]   is analytic for z in G, and  

            [Graphics:Images/IntegralRepresentationMod_gr_72.gif].  

Proof.

Demonstration for Theorem 6.11.

 

    We now generalize Theorem 6.10 to give an integral representation for the [Graphics:Images/IntegralRepresentationMod_gr_73.gif] derivative, [Graphics:Images/IntegralRepresentationMod_gr_74.gif]. We use Leibniz's rule in the proof and note that this method of proof is a mnemonic device for remembering Theorem 6.12.

 

Theorem 6.12 (Cauchy's Integral Formulae for Derivatives).  Let [Graphics:Images/IntegralRepresentationMod_gr_77.gif] be analytic in the simply connected domain D, and let C be a simple closed positively oriented contour that lies in D.  If z is a point that lies interior to C, then for any integer [Graphics:Images/IntegralRepresentationMod_gr_78.gif], we have

            
[Graphics:Images/IntegralRepresentationMod_gr_79.gif].  

Proof.

 Proof of Theorem 6.12 is in the book.
Complex Analysis for Mathematics and Engineering

 

Example 6.24.  Let  [Graphics:Images/IntegralRepresentationMod_gr_80.gif]  denote a fixed complex value.  Show that, if C is a simple closed positively oriented contour such that  [Graphics:Images/IntegralRepresentationMod_gr_81.gif]  lies interior to C, then  

               [Graphics:Images/IntegralRepresentationMod_gr_82.gif],   and
(6-50)
               [Graphics:Images/IntegralRepresentationMod_gr_83.gif],   for any integer  [Graphics:Images/IntegralRepresentationMod_gr_84.gif]

[Graphics:Images/IntegralRepresentationMod_gr_85.gif]

Solution.  We let [Graphics:Images/IntegralRepresentationMod_gr_102.gif].  Then [Graphics:Images/IntegralRepresentationMod_gr_103.gif] for [Graphics:Images/IntegralRepresentationMod_gr_104.gif].  Theorem 6.10 implies that the value of the first integral in Equations (6-50) is

 

            [Graphics:Images/IntegralRepresentationMod_gr_105.gif],  

 

and Theorem 6.12 further implies that  

 

            [Graphics:Images/IntegralRepresentationMod_gr_106.gif].  

 

This result is the same as that proven earlier in Corollary 6.1.  Obviously, though, the technique of using Theorems 6.10 and 6.12 is easier.

Explore Solution 6.24 (a).

Explore Solution 6.24 (b).

 

Example 6.25.  Show that[Graphics:Images/IntegralRepresentationMod_gr_107.gif],  where C is the circle  [Graphics:Images/IntegralRepresentationMod_gr_108.gif]  with positive orientation.

[Graphics:Images/IntegralRepresentationMod_gr_109.gif]

Solution.  If we set  [Graphics:Images/IntegralRepresentationMod_gr_110.gif],  then a straightforward calculation shows that  [Graphics:Images/IntegralRepresentationMod_gr_111.gif].   Using Cauchy's integral formulas with [Graphics:Images/IntegralRepresentationMod_gr_112.gif], we conclude that  

            [Graphics:Images/IntegralRepresentationMod_gr_113.gif]  

Explore Solution 6.25.

 

    We now state two important corollaries of Theorem 6.12.

 

Corollary 6.2.  If [Graphics:Images/IntegralRepresentationMod_gr_129.gif] is analytic in the domain D, then all derivatives  [Graphics:Images/IntegralRepresentationMod_gr_130.gif] [Graphics:Images/IntegralRepresentationMod_gr_131.gif]  exists for  [Graphics:Images/IntegralRepresentationMod_gr_132.gif]  (and therefore are analytic in D).

Proof.

Proof of Corollary 6.2 is in the book.
Complex Analysis for Mathematics and Engineering

 

Remark 6.3.  This result is interesting, as it illustrates a big difference between real and complex functions.  A real function [Graphics:Images/IntegralRepresentationMod_gr_133.gif] can have the property that [Graphics:Images/IntegralRepresentationMod_gr_134.gif] exists everywhere in a domain D, but [Graphics:Images/IntegralRepresentationMod_gr_135.gif] exists nowhere.  Corollary 6.2 states that if a complex function [Graphics:Images/IntegralRepresentationMod_gr_136.gif] has the property that [Graphics:Images/IntegralRepresentationMod_gr_137.gif]exists everywhere in a domain D, then, remarkably, all derivatives of [Graphics:Images/IntegralRepresentationMod_gr_138.gif] exist in D.  

 

Corollary 6.3.  If [Graphics:Images/IntegralRepresentationMod_gr_139.gif] is a harmonic function at each point [Graphics:Images/IntegralRepresentationMod_gr_140.gif] in the domain D, then all partial derivatives  [Graphics:Images/IntegralRepresentationMod_gr_141.gif], [Graphics:Images/IntegralRepresentationMod_gr_142.gif], [Graphics:Images/IntegralRepresentationMod_gr_143.gif], [Graphics:Images/IntegralRepresentationMod_gr_144.gif], [Graphics:Images/IntegralRepresentationMod_gr_145.gif]  exists and are harmonic functions.  

Proof.

Proof of Corollary 6.3 is in the book.
Complex Analysis for Mathematics and Engineering

 

Extra Example 1.  Show that the partial derivatives of  [Graphics:Images/IntegralRepresentationMod_gr_146.gif] are harmonic functions.  

Explore Extra Solution 1.

 

Exercises for Section 6.5.  Integral Representations for Analytic Functions

 

Library Research Experience for Undergraduates

Cauchy-Goursat Theorem  

Cauchy's Integral Formula  

 

 

 

 

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Theorems of Morera and Liouville and Extensions

 

 

 

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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2012 John H. Mathews, Russell W. Howell