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 + ... + ar^n-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 + ... + r^n-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)

  Wylie. Advanced Engineering Mathematics