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Since f ( z + ω 1 )= f ( z ), we have
ω 1
f ( z ) dz = 0
ω 2
ω 2
f ( z ) dz =
f ( z ) dz.
ω 1 + ω 2
0
Similarly,
ω 1 + ω 2
0
f ( z ) dz =
f ( z ) dz.
ω 2
ω 1
Therefore, the sum of the four integrals is 0. There is a small technicality
that we have passed over. The function f is not allowed to have any poles on
the path of integration. If it does, adjust the path with a small detour around
such points as in Figure 9.2. The integrals cancel, just as in the above. This
proves (2).
Ω 3
Ω 1
Ω 2
0
Figure 9.2
w ) r g ( z ), where g ( w ) is finite and
Suppose r =ord w f .Then f ( z )=( z
nonzero. Then
f ( z )
f ( z )
w + g ( z )
r
=
g ( z ) ,
z
so
Res w f
= r.
f
If f is doubly periodic, then f
is doubly periodic. Therefore, (2) applied to
f /f yields
Res w f
=0 .
2 πi
w
ord w f =2 πi
w
f
F
F
This proves (3).
For (4), we have
2 πi
w
Res w z f
=
ord w f =2 πi
w
z f
w
·
f dz.
f
∂F
F
F
 
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