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Prove: Laurent’s theorem. Let f(z) be analytic throughout the closed annular region R
bounded by two concentric circles, C_{1} and C_{2} centered at point a with respective radii r_{1} and r_{2} (r_{1}
> r_{2}). 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 + ar^{2} + ... + ar^{n-1} = a(1- r^{n})/(1- r) ,

an identity valid for both real and complex numbers.

Letting a = 1 in 5) we get

6) 1 + r + r^{2} + ... + r^{n-1} = 1/(1 - r) - r^{n}/(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 C_{2} in
Cauchy’s formula is clockwise and here C_{2} 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 C_{1}

where
is a constant. Second, note that on C_{1}, the function f(w) must acquire some finite
maximum value i.e. it is bounded. Why? Because C_{1} lies in an analytic region and so there can
be no singular points on C_{1} where the function might become infinite. Let M be the maximum of
f(w) on C_{1}. The radius r_{1} of C_{1} is given by r_{1} = |w - a|. We will also make use of the following
observation:

|w - z| = |(w - a) - (z - a)|
r_{1} - |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 C_{1} is 2πr_{1} so

Because < 1 it can be seen that

since

Proof.
. First note that since w is on C_{2}

where
is a constant. Second, note that on C_{2}, the function f(w) must acquire some finite
maximum value i.e. it is bounded. Let M be the maximum of f(w) on C_{2}. The radius r_{2} of C_{2} is
given by r_{2} = |w - a|. We will also make use of the following observation:

|z - w| = |(z - a) - (w - a)|
|z - a| - r_{2}

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 C_{2} is 2πr_{2} so

Because < 1 it can be seen that

since

Since f(z) is analytic throughout the annular region between C_{1} and C_{2}, by the Principle of the
Deformation of Contours, the paths of integration C_{1} and C_{2} can be replaced by any other curve C
within the region and encircling C_{1}.

Principle of the Deformation of Contours. The line integral of an analytic function around any
closed curve C_{1} is equal to the line integral of the same function around any other closed curve C_{2}
into which C_{1} 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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