Taylor and Laurent series

24.11 TAYLOR AND LAURENT SERIES
Further, it may be proved by induction that the nth derivative of f(z) is also
given by a Cauchy integral,
0
f(z) dz
n!
.
(24.48)
f (n) (z0 ) =
2πi C (z − z0 )n+1
Thus, if the value of the analytic function is known on C then not only may the
value of the function at any interior point be calculated, but also the values of
all its derivatives.
The observant reader will notice that (24.48) may also be obtained by the
formal device of differentiating under the integral sign with respect to z0 in
Cauchy’s integral formula (24.46):
0
f(z)
∂n
1
f (n) (z0 ) =
dz
2πi C ∂z0n (z − z0 )
0
f(z) dz
n!
.
=
2πi C (z − z0 )n+1
Suppose that f(z) is analytic inside and on a circle C of radius R centred on the point
z = z0 . If |f(z)| ≤ M on the circle, where M is some constant, show that
|f (n) (z0 )| ≤
From (24.48) we have
|f (n) (z0 )| =
Mn!
.
Rn
(24.49)
0
n! f(z) dz ,
n+1
2π C (z − z0 ) and on using (24.39) this becomes
n! M
Mn!
2πR =
.
2π R n+1
Rn
This result is known as Cauchy’s inequality. |f (n) (z0 )| ≤
We may use Cauchy’s inequality to prove Liouville’s theorem, which states that
if f(z) is analytic and bounded for all z then f is a constant. Setting n = 1 in
(24.49) and letting R → ∞, we find |f (z0 )| = 0 and hence f (z0 ) = 0. Since f(z) is
analytic for all z, we may take z0 as any point in the z-plane and thus f (z) = 0
for all z; this implies f(z) = constant. Liouville’s theorem may be used in turn to
prove the fundamental theorem of algebra (see exercise 24.9).
24.11 Taylor and Laurent series
Following on from (24.48), we may establish Taylor’s theorem for functions of a
complex variable. If f(z) is analytic inside and on a circle C of radius R centred
on the point z = z0 , and z is a point inside C, then
f(z) =
∞
an (z − z0 )n ,
n=0
853
(24.50)
COMPLEX VARIABLES
where an is given by f (n) (z0 )/n!. The Taylor expansion is valid inside the region
of analyticity and, for any particular z0 , can be shown to be unique.
To prove Taylor’s theorem (24.50), we note that, since f(z) is analytic inside
and on C, we may use Cauchy’s formula to write f(z) as
0
f(ξ)
1
dξ,
(24.51)
f(z) =
2πi C ξ − z
where ξ lies on C. Now we may expand the factor (ξ − z)−1 as a geometric series
in (z − z0 )/(ξ − z0 ),
n
∞ 1
1 z − z0
=
,
ξ−z
ξ − z0
ξ − z0
n=0
so (24.51) becomes
∞ f(ξ) z − z0 n
dξ
ξ − z0
C ξ − z0 n=0
0
∞
f(ξ)
1 =
(z − z0 )n
dξ
n+1
2πi
C (ξ − z0 )
f(z) =
1
2πi
0
n=0
∞
2πif (n) (z0 )
1 ,
(z − z0 )n
=
2πi
n!
(24.52)
n=0
where we have used Cauchy’s integral formula (24.48) for the derivatives of
f(z). Cancelling the factors of 2πi, we thus establish the result (24.50) with
an = f (n) (z0 )/n!.
Show that if f(z) and g(z) are analytic in some region R, and f(z) = g(z) within some
subregion S of R, then f(z) = g(z) throughout R.
It is simpler to consider the (analytic) function h(z) = f(z) − g(z), and to show that because
h(z) = 0 in S it follows that h(z) = 0 throughout R.
If we choose a point z = z0 in S, then we can expand h(z) in a Taylor series about z0 ,
h(z) = h(z0 ) + h (z0 )(z − z0 ) + 12 h (z0 )(z − z0 )2 + · · · ,
which will converge inside some circle C that extends at least as far as the nearest part of
the boundary of R, since h(z) is analytic in R. But since z0 lies in S, we have
h(z0 ) = h (z0 ) = h (z0 ) = · · · = 0,
and so h(z) = 0 inside C. We may now expand about a new point, which can lie anywhere
within C, and repeat the process. By continuing this procedure we may show that h(z) = 0
throughout R.
This result is called the identity theorem and, in fact, the equality of f(z) and g(z)
throughout R follows from their equality along any curve of non-zero length in R, or even
at a countably infinite number of points in R. So far we have assumed that f(z) is analytic inside and on the (circular)
contour C. If, however, f(z) has a singularity inside C at the point z = z0 , then it
cannot be expanded in a Taylor series. Nevertheless, suppose that f(z) has a pole
854
24.11 TAYLOR AND LAURENT SERIES
of order p at z = z0 but is analytic at every other point inside and on C. Then
the function g(z) = (z − z0 )p f(z) is analytic at z = z0 , and so may be expanded
as a Taylor series about z = z0 :
g(z) =
∞
bn (z − z0 )n .
(24.53)
n=0
Thus, for all z inside C, f(z) will have a power series representation of the form
a−1
a−p
+ ··· +
+ a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · ,
f(z) =
(z − z0 )p
z − z0
(24.54)
with a−p = 0. Such a series, which is an extension of the Taylor expansion, is
called a Laurent series. By comparing the coefficients in (24.53) and (24.54), we
see that an = bn+p . Now, the coefficients bn in the Taylor expansion of g(z) are
seen from (24.52) to be given by
0
1
g (n) (z0 )
g(z)
=
dz,
bn =
n!
2πi
(z − z0 )n+1
and so for the coefficients an in (24.54) we have
0
0
g(z)
f(z)
1
1
an =
dz
=
dz,
2πi
(z − z0 )n+1+p
2πi
(z − z0 )n+1
an expression that is valid for both positive and negative n.
The terms in the Laurent series with n ≥ 0 are collectively called the analytic
part, whilst the remainder of the series, consisting of terms in inverse powers of
z − z0 , is called the principal part. Depending on the nature of the point z = z0 ,
the principal part may contain an infinite number of terms, so that
f(z) =
+∞
an (z − z0 )n .
(24.55)
n=−∞
In this case we would expect the principal part to converge only for |(z − z0 )−1 |
less than some constant, i.e. outside some circle centred on z0 . However, the
analytic part will converge inside some (different) circle also centred on z0 . If the
latter circle has the greater radius then the Laurent series will converge in the
region R between the two circles (see figure 24.12); otherwise it does not converge
at all.
In fact, it may be shown that any function f(z) that is analytic in a region
R between two such circles C1 and C2 centred on z = z0 can be expressed as
a Laurent series about z0 that converges in R. We note that, depending on the
nature of the point z = z0 , the inner circle may be a point (when the principal
part contains only a finite number of terms) and the outer circle may have an
infinite radius.
We may use the Laurent series of a function f(z) about any point z = z0 to
855
COMPLEX VARIABLES
y
R
C2
C1
z0
x
Figure 24.12 The region of convergence R for a Laurent series of f(z) about
a point z = z0 where f(z) has a singularity.
classify the nature of that point. If f(z) is actually analytic at z = z0 , then in
(24.55) all an for n < 0 must be zero. It may happen that not only are all an
zero for n < 0 but a0 , a1 , . . . , am−1 are all zero as well. In this case, the first
non-vanishing term in (24.55) is am (z − z0 )m , with m > 0, and f(z) is then said to
have a zero of order m at z = z0 .
If f(z) is not analytic at z = z0 , then two cases arise, as discussed above (p is
here taken as positive):
(i) it is possible to find an integer p such that a−p = 0 but a−p−k = 0 for all
integers k > 0;
(ii) it is not possible to find such a lowest value of −p.
In case (i), f(z) is of the form (24.54) and is described as having a pole of order
p at z = z0 ; the value of a−1 (not a−p ) is called the residue of f(z) at the pole
z = z0 , and will play an important part in later applications.
For case (ii), in which the negatively decreasing powers of z − z0 do not
terminate, f(z) is said to have an essential singularity. These definitions should be
compared with those given in section 24.6.
Find the Laurent series of
f(z) =
1
z(z − 2)3
about the singularities z = 0 and z = 2 (separately). Hence verify that z = 0 is a pole of
order 1 and z = 2 is a pole of order 3, and find the residue of f(z) at each pole.
To obtain the Laurent series about z = 0, we make the factor in parentheses in the
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24.11 TAYLOR AND LAURENT SERIES
denominator take the form (1 − αz), where α is some constant, and thus obtain
1
8z(1 − z/2)3
z (−3)(−4) z 2 (−3)(−4)(−5) z 3
1
+
−
−
=−
+
+ ···
1 + (−3) −
8z
2
2!
2
3!
2
f(z) = −
1
3
3z
5z 2
−
−
−
− ··· .
8z
16 16
32
Since the lowest power of z is −1, the point z = 0 is a pole of order 1. The residue of f(z)
at z = 0 is simply the coefficient of z −1 in the Laurent expansion about that point and is
equal to −1/8.
The Laurent series about z = 2 is most easily found by letting z = 2 + ξ (or z − 2 = ξ)
and substituting into the expression for f(z) to obtain
=−
1
1
= 3
(2 + ξ)ξ 3
2ξ (1 + ξ/2)
2 3 4
1
ξ
ξ
ξ
ξ
+
= 3 1−
−
+
− ···
2ξ
2
2
2
2
f(z) =
1
1
1
1
ξ
− 2 +
−
+
− ···
2ξ 3
4ξ
8ξ
16 32
1
1
1
z−2
1
−
+
−
+
− ··· .
=
2(z − 2)3
4(z − 2)2
8(z − 2) 16
32
=
From this series we see that z = 2 is a pole of order 3 and that the residue of f(z) at z = 2
is 1/8. As we shall see in the next few sections, finding the residue of a function
at a singularity is of crucial importance in the evaluation of complex integrals.
Specifically, formulae exist for calculating the residue of a function at a particular
(singular) point z = z0 without having to expand the function explicitly as a
Laurent series about z0 and identify the coefficient of (z − z0 )−1 . The type of
formula generally depends on the nature of the singularity at which the residue
is required.
Suppose that f(z) has a pole of order m at the point z = z0 . By considering the Laurent
series of f(z) about z0 , derive a general expression for the residue R(z0 ) of f(z) at z = z0 .
Hence evaluate the residue of the function
f(z) =
exp iz
(z 2 + 1)2
at the point z = i.
If f(z) has a pole of order m at z = z0 , then its Laurent series about this point has the
form
a−1
a−m
+ a0 + a1 (z − z0 ) + a2 (z − z0 )2 + · · · ,
+ ··· +
f(z) =
(z − z0 )m
(z − z0 )
which, on multiplying both sides of the equation by (z − z0 )m , gives
(z − z0 )m f(z) = a−m + a−m+1 (z − z0 ) + · · · + a−1 (z − z0 )m−1 + · · · .
857