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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

where

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:

Proof. First we write

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

We now wish to employ the following identity:

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)

We now substitute 7) into 3) to obtain

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

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

where

and

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

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

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

where

and

From 9), 10) and 12) we have

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

Proof. . First note that since w is on C1

where 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)| r1 - |z - a|

We now employ Property 5 of line integrals which states that

where |f(z)| 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

Because < 1 it can be seen that

since

Proof. . First note that since w is on C2

where 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)| |z - a| - r2

We now employ Property 5 of line integrals which states that

where |f(z)| 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

Because < 1 it can be seen that

since

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)