Cryptography Reference
In-Depth Information
Figure 9.4
imaginary part
N
,where
N
is large enough that
f
(
z
) has no zeros or poles
with imaginary part greater than
N
, except perhaps at
i∞
. Thisispossible
since
f
(
z
)=
q
n
(
a
n
+
a
n
+1
q
+
···
)
.
Since the series
a
n
+
a
n
+1
q
+
···
is assumed to converge for
q
small, it is finite
and is approximately equal to
a
n
= 0 for suciently small
q
.
As in the proof of Theorem 9.1, we have
dz
=
z∈F
z
f
(
z
)
f
(
z
)
1
2
πi
ord
z
(
f
)
.
C
=
i,ρ
Since
11
∈ SL
2
(
Z
)givesthemap
z → z
+1, we have
01
f
(
z
)=
f
(
z
+1)
.
(9.16)
Therefore, the integrals over the left and right vertical parts of
C
are the
same, except that they are in opposite directions, so they cancel each other.
Now we'll show that the integral over the part of the unit circle to the left of
i
cancels the part to the right. This is proved by using the fact that
0
1
10
−
∈
SL
2
(
Z
)givesthemap
z →−
1
/z
, which interchanges the left and right arcs of
the unit circle. In addition, differentiating the relation
f
(
−
1
/z
)=
f
(
z
) yields
−
1
z
d
−
1
z
=
f
f
f
f
(
z
)
dz.
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