Cryptography Reference
In-Depth Information
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
Search WWH ::
Custom Search