Page 129 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Then by cor. to Cauchy’s Theorem, we have

n
 f(z) dz =  f(z) dz (4)
C r 1 r 

Now, if a is a pole of order m, then by Laurent’s theorem, f(z) can be expressed as
r
m b
f(z) = (z) +  s
s 1 (z a ) r s

where (z) is regular within and on  .
r
Then
m b
 f(z) dz =   s s dz
 
 r s 1 (z a ) r
r
(5)
where  f(z) dz  0, by Cauchy’s theorem
 r
Now, on  |z-a | =  i.e. z = a + e i
r
r
r
 dz =  ie d
i
where  varies from 0 to 2 as the point z moves once round g . r
m 2
Thus,  f(z) dz =  b  1-s 0  e (1 s)i id

s

 s 1
r
= 2pi b 1
= 2pi [Residue of f(z) at a ]
r

where 0  2 e (1 s)i d =  0, if s  1 1

2 if s 


Hence, from (4), we find
n
 f(z)dz =  2 i [Residue of f(z) at a ]

r
C r 1

n
= 2i   Residue of f(z) at a r  
 r 1 

= 2i [sum of Residues of f(z) at its poles inside C.]
which proves the theorem.

Remark. If f(z) can be expressed in the form f(z) =  (z) where (z) is analytic and (a)  0, then
(z a) m

the pole z = a is a pole of type I or overt.
If f(z) is of the form f(z) =  (z) , where (z) and (z) are analytic and (a)  0 and (z) has a zero
 (z)
of order m at z = a, then z = a is a pole of type II or covert. Actually, whether a pole of f(z) is overt
or covert, is a matter of how f(z) is written.

122 LOVELY PROFESSIONAL UNIVERSITY