Residue theorem

COMPLEX VARIABLES
Differentiating both sides m − 1 times, we obtain
dm−1
[(z − z0 )m f(z)] = (m − 1)! a−1 +
bn (z − z0 )n ,
dz m−1
n=1
∞
for some coefficients bn . In the limit z → z0 , however, the terms in the sum disappear, and
after rearranging we obtain the formula
dm−1
1
R(z0 ) = a−1 = lim
[(z − z0 )m f(z)] ,
(24.56)
m−1
z→z0
(m − 1)! dz
which gives the value of the residue of f(z) at the point z = z0 .
If we now consider the function
exp iz
exp iz
f(z) = 2
=
,
(z + 1)2
(z + i)2 (z − i)2
we see immediately that it has poles of order 2 (double poles) at z = i and z = −i. To
calculate the residue at (for example) z = i, we may apply the formula (24.56) with m = 2.
Performing the required differentiation, we obtain
d
d
exp iz
[(z − i)2 f(z)] =
dz
dz (z + i)2
1
=
[(z + i)2 i exp iz − 2(exp iz)(z + i)].
(z + i)4
Setting z = i, we find the residue is given by
i
1 1 R(i) =
−4ie−1 − 4ie−1 = − . 1! 16
2e
An important special case of (24.56) occurs when f(z) has a simple pole (a pole
of order 1) at z = z0 . Then the residue at z0 is given by
R(z0 ) = lim [(z − z0 )f(z)] .
z→z0
(24.57)
If f(z) has a simple pole at z = z0 and, as is often the case, has the form
g(z)/h(z), where g(z) is analytic and non-zero at z0 and h(z0 ) = 0, then (24.57)
becomes
(z − z0 )g(z)
(z − z0 )
= g(z0 ) lim
R(z0 ) = lim
z→z0
z→z0
h(z)
h(z)
1
g(z0 )
= ,
(24.58)
= g(z0 ) lim z→z0 h (z)
h (z0 )
where we have used l’Hôpital’s rule. This result often provides the simplest way
of determining the residue at a simple pole.
24.12 Residue theorem
Having seen from Cauchy’s theorem that the value of an integral round a closed
contour C is zero if the integrand is analytic inside the contour, it is natural to
ask what value it takes when the integrand is not analytic inside C. The answer
to this is contained in the residue theorem, which we now discuss.
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24.12 RESIDUE THEOREM
Suppose the function f(z) has a pole of order m at the point z = z0 , and so
can be written as a Laurent series about z0 of the form
f(z) =
∞
an (z − z0 )n .
(24.59)
n=−m
Now consider the integral I of f(z) around a closed contour C that encloses
z = z0 , but no other singular points. Using Cauchy’s theorem, this integral has
the same value as the integral around a circle γ of radius ρ centred on z = z0 ,
since f(z) is analytic in the region between C and γ. On the circle we have
z = z0 + ρ exp iθ (and dz = iρ exp iθ dθ), and so
0
I = f(z) dz
γ
=
=
∞
n=−m
∞
0
(z − z0 )n dz
an
2π
iρn+1 exp[i(n + 1)θ] dθ.
an
0
n=−m
For every term in the series with n = −1, we have
n+1
2π
2π
iρ exp[i(n + 1)θ]
iρn+1 exp[i(n + 1)θ] dθ =
= 0,
i(n + 1)
0
0
but for the n = −1 term we obtain
2π
i dθ = 2πi.
0
Therefore only the term in (z − z0 )−1 contributes to the value of the integral
around γ (and therefore C), and I takes the value
0
I=
f(z) dz = 2πia−1 .
(24.60)
C
Thus the integral around any closed contour containing a single pole of general
order m (or, by extension, an essential singularity) is equal to 2πi times the residue
of f(z) at z = z0 .
If we extend the above argument to the case where f(z) is continuous within
and on a closed contour C and analytic, except for a finite number of poles,
within C, then we arrive at the residue theorem
0
f(z) dz = 2πi
Rj ,
(24.61)
C
j
where j Rj is the sum of the residues of f(z) at its poles within C.
The method of proof is indicated by figure 24.13, in which (a) shows the original
contour C referred to in (24.61) and (b) shows a contour C giving the same value
859
COMPLEX VARIABLES
C
C
C
(b)
(a)
Figure 24.13 The contours used to prove the residue theorem: (a) the original
contour; (b) the contracted contour encircling each of the poles.
to the integral, because f is analytic between C and C . Now the contribution to
the C integral from the polygon (a triangle for the case illustrated) joining the
small circles is zero, since f is also analytic inside C . Hence the whole value of
the integral comes from the circles and, by result (24.60), each of these contributes
2πi times the residue at the pole it encloses. All the circles are traversed in their
positive sense if C is thus traversed and so the residue theorem follows. Formally,
Cauchy’s theorem (24.40) is a particular case of (24.61) in which C encloses no
poles.
Finally we prove another important result, for later use. Suppose that f(z) has
a simple pole at z = z0 and so may be expanded as the Laurent series
f(z) = φ(z) + a−1 (z − z0 )−1 ,
where φ(z) is analytic within some neighbourhood surrounding z0 . We wish to
find an expression for the integral I of f(z) along an open contour C, which is
the arc of a circle of radius ρ centred on z = z0 given by
|z − z0 | = ρ,
θ1 ≤ arg(z − z0 ) ≤ θ2 ,
(24.62)
where ρ is chosen small enough that no singularity of f, other than z = z0 , lies
within the circle. Then I is given by
f(z) dz =
φ(z) dz + a−1 (z − z0 )−1 dz.
I=
C
C
C
If the radius of the arc C is now allowed to tend to zero, then the first integral
tends to zero, since the path becomes of zero length and φ is analytic and
therefore continuous along it. On C, z = ρeiθ and hence the required expression
for I is
θ2
1
iθ
f(z) dz = lim a−1
iρe
dθ
= ia−1 (θ2 − θ1 ). (24.63)
I = lim
iθ
ρ→0 C
ρ→0
θ1 ρe
860