5.3 Laurent Series

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You may have noticed a certain interest in functions which are reciprocals

of polynomials. The reason of course is that they are easy to compute,

just as polynomials are. It is worth looking at functions which are ratios of

polynomials also, and indeed functions which are ratios of other functions we

already know. We shall come back to this later, but for the moment consider

a function such as

ez

z2

This is a ratio of analytic functions and is hence analytic except at the zeros

of the denominator. There are two roots, both the same, so we have a

singularity at z = 0. We can divide out the power series to get

ez

z2 _

1

z2

+

1

z

+

1

2!

+

z

3!

+ _ _ _

I haven't wanted to say these are equal: this would beg the question. But

the question more or less raises itself, are these equal? Or to put it another

way, does the expression on the right converge if jzj > 0, and if so, does it

converge to ez=z2?

The answer is 'yes' to both parts.

More generally, suppose we have a function which is analytic in the annulus

r < jz􀀀cj < R, for some point c, the centre of the annulus. Then it will in

general have an expansion in terms of integral powers, some or all of which

may be negative. This is called a Laurent Series for the function.

More formally:

De_nition 5.3.1 (Laurent Series)

5.3. LAURENT SERIES 139

For any w, the integer power series

1 X

􀀀1

ak(z 􀀀 c)k; k 2 Z

is de_ned to be

1 X

0

ak(z 􀀀 c)k +

1 X

1

a

􀀀k(

1

z 􀀀 c

)k

when both of these series converge.

Theorem 5.3 (Laurent's Theorem)

If f is analytic on the annulus r < jz 􀀀 cj < R, for some point c, then f is

equal to the Laurent series

f(z) =

1 X

􀀀1

ak(z 􀀀 c)k

where the coe_cients ak can be computed from:

ak =

1

2_i IC

f(z)

(z 􀀀 c)1+k dz

if k is positive or zero, and

ak =

1

2_i IC

f(z)

(z 􀀀 c)1􀀀k dz

when k is negative. C is any simple closed loop around the centre c which is

contained in the annulus and goes in the positive sense.

No Proof: See Mathews and Howell or any standard text. 2

Exercise 5.3.1 Try showing a similar result for the Taylor series for an analytic

function; i.e. try to get an expression for the Taylor series coe_cients

in terms of path integrals.

It is the case that Laurent series about any point, like Power series, are

unique when they converge.

The following result is extremely useful:

140 CHAPTER 5. TAYLOR AND LAURENT SERIES

Theorem 5.4 (Di_erentiability of Laurent Series)

The Laurent series for a function analytic in an annulus if di_erentiated

termwise gives the derivative of the function.

No Proof: 2

Since the case where all the negative coe_cients are zero reduces to the case

of the Taylor series, this is also true for Taylor Series. It is not generally

true that if a function is given by a sequence of approximating functions, the

derivative is given by the sequence of derivatives. After all,

1

n

sin nx

gets closer and closer to the zero function as n increases. But the derivatives

cos nx certainly do not get closer to anything.

This tells us yet again that the analytic functions are very special and that

they behave in particularly pleasant ways, all things considered.

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