Complex Analysis - Maths KU

364 Chapter 6 Series and Residues
f(z) =(z −z0) n φ(z), where φ is analytic at z0 and φ(z0) �= 0.We differentiate
f by the product rule,
f ′ (z) =(z − z0) n φ ′ (z)+n(z − z0) n−1 φ(z),
and divide this expression by f.In some punctured disk centered at z0, we
have
f ′ (z)
f(z) = (z − z0) nφ ′ (z)+n(z − z0) n−1φ(z) (z − z0) n =
φ(z)
φ′ (z) n
+ . (29)
φ(z) z − z0
The result in (29) shows that the integrand f ′ (z)/f(z) has a simple pole at
z0 and the residue at that point is
� ′ f (z)
Res
f(z) ,z0
�
� �
′ φ (z) n
= lim (z − z0) +
z→z0 φ(z) z − z0
�
= lim (z − z0)
z→z0
φ′ �
(z)
+ n =0+n = n,
φ(z)
(30)
which is the order of the zero z0.
Now if zp is a pole of order m of f within C, then by (7) of Section 6.4
we can write f(z) =g(z)/(z − zp) m , where g is analytic at zp and g(zp) �= 0.
By differentiating, in this case f(z) =(z − zp) −m g(z), we have
f ′ (z) =(z − zp) −m g ′ (z) − m(z − zp) −m−1 g(z).
Therefore, in some punctured disk centered at zp,
f ′ (z)
f(z) = (z − zp) −mg ′ (z) − m(z − zp) −m−1g(z) (z − zp) −m =
g(z)
g′ (z) −m
+ . (31)
g(z) z − zp
We see from (31) that the integrand f ′ (z)/f(z) has a simple pole at zp.Proceeding
as in (30), we also see that the residue at zp is equal to −m, which is
the negative of the order of the pole of f.
Finally, suppose that z01 ,z02 ,...,z0r and zp1 ,zp2 ,...,zps are the
zeros and poles of f within C and suppose further that the order of the
zeros are n1,n2, ...,nrand that order of the poles are m1, m2, ...,ms.
Then each of these points is a simple pole of the integrand f ′ (z)/f(z) with
corresponding residues n1,n2, ...,nrand −m1, −m2, ...,−ms.It follows
from the residue theorem (Theorem 6.16) that �
C f ′ (z) dz/f(z) is equal to
2πi times the sum of the residues at the poles:
�
f
C
′ �
r� � ′ (z)
f (z)
dz =2πi Res
f(z) f(z)
k=1
,z0k
� s�
� ′ f (z)
+ Res
f(z)
k=1
,zpk
�� �
r� s�
�
=2πi nk + (−mk) =2πi[N0 − Np].
k=1
k=1
Dividing by 2πi establishes (28). ✎