5.4 Some Sums

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The series

1

1 + z

= 1􀀀z +z2

􀀀z3 +_ _ _ =

1 X

0

(􀀀1)nzn

converges if jzj < 1. This is 'Well Known fact' number 137 or thereabouts.

We can get a Laurent Series for this as follows: _nd a series for 1=(1 + 1=w)

by the usual trick of doing (very) long division to get

1

1 + 1

w

= w 􀀀w2 +w3

􀀀w4 +_ _ _ =

1 X

1

(􀀀1)n+1wn

and then put z = 1=w to get:

1

1 + z

=

1

z 􀀀

1

z2 +

1

z3 􀀀

1

z4 +_ _ _ =

1 X

1

(􀀀1)n+1 1

zn

5.4. SOME SUMS 141

This converges for jzj > 1. It clearly goes bung when z = 􀀀1, and equally

clearly is a geometric series with ratio less than 1 provided jzj > 1.

There are therefore two di_erent Laurent series for 1

1+z , one inside the unit

disk, one outside. One is actually a Taylor Series, which is just a special case.

Suppose we have a function like

1

1 􀀀 z

+

1

z 􀀀 2i

=

1 􀀀 2i

z2 􀀀 (1 + 2i)z + 2i

This has singularities at z = 1 and z􀀀2i, where the modulus of the function

goes through the roof.

The function can be expanded about the origin to get:

1 + z +z2 +z3 +_ _ _+

i

2 􀀀

z

4

+

z2

8

+_ _ _

which converges inside the unit disk.

In the annulus given by 1 < jzj < 2 it can be written as

􀀀

1

z 􀀀

1

z2 􀀀

1

z3 􀀀_ _ _+

i

2 􀀀

z

4

+

z2

8

+_ _ _

And in the annulus jzj > 2 it can be written:

􀀀

1

z 􀀀

1

z2 􀀀

1

z3 􀀀_ _ _+

1

z

+

2i

z2 􀀀

4

z3 􀀀

8i

z4

+

16

z5

+ _ _ _

Exercise 5.4.1 Con_rm the above or _nd my error.

Substitutions for terms are valid providing care is taken about the radius of

convergence of both series.

Laurent series expansions about the origin have been produced by some simple

division. The uniqueness of the Laurent expansions tells us that these

have to be the right answers. Next we consider some simple tricks for getting

expansions about other points:

Example 5.4.1 Find a Laurent expansion of 1

1+z about i

142 CHAPTER 5. TAYLOR AND LAURENT SERIES

We write

1

1 + z

=

1

(1 + i) + (z 􀀀i)

= (

1

1 + i

)(

1

1 + z􀀀i

1+i

)

Then we have the Taylor expansion

1

1 + z􀀀i

1+i

= 1􀀀

z 􀀀i

1 + i

+

(z 􀀀i)2

(1 + i)2 􀀀

(z 􀀀 i)3

(1 + i)3

+ _ _ _

valid for j

z􀀀i

1+i j < 1 i.e. for jz 􀀀 ij < p2. We also have

1

1 + z􀀀i

1+i

=

1 + i

z 􀀀i 􀀀

(1 + i)2

(z 􀀀 i)2

+ _ _ _

valid when jz 􀀀 ij > p2.

Example 5.4.2 Find a Laurent expansion for

(1 􀀀 z)3

z 􀀀 2

about 1.

(1 􀀀 z)3

z 􀀀 2

=

(z 􀀀 1)3

2 􀀀 z

=

(z 􀀀 1)3

1 􀀀 (z 􀀀 1)

Putting w = z 􀀀 1 we get

w3 1

1 􀀀 w

= w3(􀀀

1

w 􀀀

1

w2 􀀀

1

w3 􀀀_ _ _)

To give the _nal result

􀀀((z 􀀀 1)2 + (z 􀀀1) + 1 +

1

z 􀀀1

+

1

(z 􀀀1)2 + _ _ _)

which is valid for jz 􀀀 1j > 1.

A small amount of ingenuity may be required to beat the expressions into

the correct shape; practice does it.

5.5. POLES AND ZEROS 143

Exercise 5.4.2 Find the Laurent expansion for

1 􀀀 z

z 􀀀 3

about 1, valid for jz 􀀀 1j > 2.

Exercise 5.4.3 Make up a problem of this type and solve it.

Exercise 5.4.4 Go through the exercises on page 230, 232 of Mathews and

Howell.

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