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Prove: Laurent’s theorem. Let f(z) be analytic throughout the closed annular region R bounded by two concentric circles, C1 and C2 centered at point a with respective radii r1 and r2 (r1 > r2). Let z be a point in R. Then f(z) can be represented by


ole.gif


                                                 ole1.gif  



where



ole2.gif

ole3.gif                                                              

                                                            


and each integral is taken in the counterclockwise direction around any closed curve C in the annular region that encircles the inner boundary.


Proof. Our first step is to prove the following lemma. 


Lemma. For any three variables w, a and z the following identity holds:


ole4.gif


                                                 ole5.gif  


Proof. First we write


ole6.gif  

Now multiply both numerator and denominator of the right member by the fraction 1/(w - a) to get


ole7.gif

We now wish to employ the following identity:


ole8.gif


Where does this identity come from? It comes from the formula for the sum of a geometric progression

 

5)        a + ar + ar2 + ... + arn-1 = a(1- rn)/(1- r) ,


an identity valid for both real and complex numbers.


Letting a = 1 in 5) we get

 

6)        1 + r + r2 + ... + rn-1 = 1/(1 - r) - rn/(1- r)

 

which, with a transposition, is 4).


Now let r = (z - a)/(w-a) and substitute into 4)



ole9.gif


                                                             ole10.gif


We now substitute 7) into 3) to obtain



ole11.gif


                                                 ole12.gif  


which is 1).


End of Proof.


Using the above lemma we will now proceed to prove our main theorem.


By Cauchy’s integral formula we have


ole13.gif


 

where the second term has a minus sign because the positive direction for traversing curve C2 in Cauchy’s formula is clockwise and here C2 is being traversed in the counterclockwise direction so we reverse the sign.


We first direct our attention to the first integral in 8). Substituting 1) above into the first integral of 8) we get



ole14.gif



                                                             ole15.gif


 

ole16.gif


where


 

ole17.gif


and

             ole18.gif


We now consider the second integral in 8). Let us rewrite 8) as follows:



ole19.gif  


where we have transferred the minus sign into the integral, interchanging z and w in the denominator. We now want an expression for 1/(z - w) analogous to the one for 1/(w - z) in 1) above. This is obtained by simply interchanging z and w in the identity. Thus for 1/(z - w) identity 1) becomes


ole20.gif


                                                 ole21.gif  


Substituting 11) into the second integral in 10) we get



ole22.gif


                                                 ole23.gif


                                     ole24.gif


where


ole25.gif


and


             ole26.gif


From 9), 10) and 12) we have


ole27.gif


                                                 ole28.gif


We now wish to show that (a) ole29.gif and (b) ole30.gif .


Proof. ole31.gif . First note that since w is on C1


             ole32.gif


where ole33.gif is a constant. Second, note that on C1, the function f(w) must acquire some finite maximum value i.e. it is bounded. Why? Because C1 lies in an analytic region and so there can be no singular points on C1 where the function might become infinite. Let M be the maximum of f(w) on C1. The radius r1 of C1 is given by r1 = |w - a|. We will also make use of the following observation:


            |w - z| = |(w - a) - (z - a)| ole34.gif r1 - |z - a|



We now employ Property 5 of line integrals which states that

 

             ole35.gif


where |f(z)| ole36.gif M ( i.e. M is an upper bound of |f(z)| on C) and L is the length of C. The length of C1 is 2πr1 so


ole37.gif


Because ole38.gif < 1 it can be seen that


             ole39.gif


since


             ole40.gif


Proof. ole41.gif . First note that since w is on C2


             ole42.gif


where ole43.gif is a constant. Second, note that on C2, the function f(w) must acquire some finite maximum value i.e. it is bounded. Let M be the maximum of f(w) on C2. The radius r2 of C2 is given by r2 = |w - a|. We will also make use of the following observation:


            |z - w| = |(z - a) - (w - a)| ole44.gif |z - a| - r2



We now employ Property 5 of line integrals which states that

 

             ole45.gif


where |f(z)| ole46.gif M ( i.e. M is an upper bound of |f(z)| on C) and L is the length of C. The length of C2 is 2πr2 so


ole47.gif


Because ole48.gif < 1 it can be seen that


             ole49.gif


since


             ole50.gif


Since f(z) is analytic throughout the annular region between C1 and C2, by the Principle of the Deformation of Contours, the paths of integration C1 and C2 can be replaced by any other curve C within the region and encircling C1.


Principle of the Deformation of Contours. The line integral of an analytic function around any closed curve C1 is equal to the line integral of the same function around any other closed curve C2 into which C1 can be continuously deformed without passing through a point where f(z) is nonanalytic.


References

  Spiegel. Complex Variables (Schaum)

  Wylie. Advanced Engineering Mathematics


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