1 Taylor and Laurent series

Many of the results in the area of series of real variables can be extended into complex variables. As an example, the concept of radius of convergence of a series is extended to the concept of a circle of convergence . If the circle of convergence of a series of complex numbers is z − z 0 = ρ then the series will converge if z − z 0 < ρ .

For example, consider the function

f ( z ) = 1 1 − z

It has a singularity at z = 1 . We can obtain the Maclaurin series, centered at z = 0 , as

f ( z ) = 1 + z + z 2 + z 3 + …

The circle of convergence is z = 1 .

The radius of convergence for a series centred on z = z 0 is the distance between z 0 and the nearest singularity.

1.1 Laurent series

One of the shortcomings of Taylor series is that the circle of convergence is often only a part of the region in which f ( z ) is analytic.

As an example, the series

1 + z + z 2 + z 3 + … converges to f ( z ) = 1 1 − z

only inside the circle z = 1 even though f ( z ) is analytic everywhere except at z = 1 .

The Laurent series is an attempt to represent f ( z ) as a series over as large a region as possible. We expand the series around a point of singularity up to, but not including, the singularity itself.

Figure 16 shows an annulus of convergence r 1 < z − z 0 < r 2 within which the Laurent series (which is an extension of the Taylor series) will converge. The extension includes negative powers of ( z − z 0 ) .

Figure 16

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Now, we state Laurent’s theorem in Key Point 4.

Key Point 4

Laurent’s Theorem

If f ( z ) is analytic through a closed annulus D centred at z = z 0 then at any point z inside D we can write

f ( z ) = a 0 + a 1 ( z − z 0 ) + a 2 ( z − z 0 ) 2 + … + b 1 ( z − z 0 ) − 1 + b 2 ( z − z 0 ) − 2 + …

where the coefficients a n and b n (for each n ) is given by

a n = 1 2 π i ∮ C f ( z ) ( z − z 0 ) n + 1 d z , b n = 1 2 π i ∮ C f ( z ) ( z − z 0 ) 1 − n d z ,

the integral being taken around any simple closed path C lying inside D and encircling the inner boundary. (Refer to Figure 16.)

Example 15

Expand f ( z ) = 1 1 − z in terms of negative powers of z which will be valid if z > 1 .

Solution

First note that 1 − z = − z 1 − 1 z so that

f ( z ) = − 1 z 1 − 1 z = − 1 z 1 + 1 z + 1 z 2 + 1 z 3 = − 1 z − 1 z 2 − 1 z 3 − 1 z 4 − …

This is valid for 1 z < 1 , that is, 1 z < 1 or z > 1 . Note that we used a binomial expansion rather than the theorem itself. Also note that together with the earlier result we are now able to expand f ( z ) = 1 1 − z everywhere, except for z = 1 .

Task!

This Task concerns f ( z ) = 1 1 + z .

  1. Using the binomial series, expand f ( z ) in terms of non-negative power of z :

    f ( z ) = ( 1 + z ) − 1 = 1 − z + z 2 − z 3 + …

  2. State the values of z for which this expansion is valid:

    z < 1 (standard result for a GP).

  3. Using the identity 1 + z = z 1 + 1 z expand f ( z ) = 1 1 + z in terms of negative powers of z and state the values of z for which your expansion is valid:

    f ( z ) = 1 z 1 + 1 z = 1 z 1 + 1 z − 1 = 1 z 1 − 1 z + 1 z 2 − 1 z 3 + … = 1 z − 1 z 2 + 1 z 3 − 1 z 4 + … Valid for 1 z < 1 i.e. z > 1 .