Cryptography Reference
In-Depth Information
However, in this case, the function
zf
/f
is not doubly periodic. The integral
may be written as a sum of four integrals, as in the proof of (2). The double
periodicity of
f
and
f
yield
ω
1
f
(
z
)
dz
=
0
z
f
(
z
)
(
z
+
ω
1
)
f
(
z
)
f
(
z
)
dz
ω
1
+
ω
2
ω
2
ω
2
f
(
z
)
dz − ω
1
ω
2
z
f
(
z
)
f
(
z
)
f
(
z
)
dz.
=
−
0
0
But
ω
2
f
(
z
)
f
(
z
)
dz
1
2
πi
0
is the winding number around 0 of the path
z
=
f
(
tω
2
)
,
0
≤
t
≤
1
.
Since
f
(0) =
f
(
ω
2
), this is a closed path. The winding number is an integer,
so
ω
2
f
(
z
)
dz
+
ω
1
z
f
(
z
)
z
f
(
z
)
f
(
z
)
dz
0
ω
1
+
ω
2
=
−ω
1
ω
2
0
f
(
z
)
f
(
z
)
dz ∈
2
πi
Z
ω
1
.
Similarly,
ω
1
+
ω
2
f
(
z
)
dz
+
0
z
f
(
z
)
z
f
(
z
)
f
(
z
)
dz
∈
2
πi
Z
ω
2
.
ω
2
ω
1
Therefore,
2
πi
w
w
·
ord
w
f
∈
2
πiL.
∈
F
This proves (4).
To prove (5), let
z
0
∈
z
0
is a doubly periodic
function whose poles are the same as the poles of
f
.By(3 ,thenumber
of zeros of
h
(
z
)in
F
(counting multiplicities) equals the number of poles
(counting multiplicities) of
h
,whichis
n
.Thisproves(5).
For (6), suppose
f
has only a simple pole, say at
w
, and no others. Then
Res
w
f
= 0 (otherwise, the pole doesn't exist). The sum in (2) has only one
term, and it is nonzero. This is impossible, so we conclude that either the
pole cannot be simple or there must be other poles.
C
.Then
h
(
z
)=
f
(
z
)
−
REMARK 9.2
As we saw in the proof of (5), part (3) says that the number
of zeros of a doubly periodic function equals the number of poles. This is a
general fact for compact Riemann surfaces (such as a torus) and for projective
algebraic curves (see [42, Ch. 8, Prop. 1] or [49, II, Cor. 6.10]).
Search WWH ::
Custom Search