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# Elliptic functions: Introduction course

Authors:
Elliptic functions:
Introduction course
Department of Mathematics, Royal Institute of Technology
Lindstedtsv¨agen 25, 10044 Stockholm, Sweden
email: tkatchev@math.kth.se
URL: http://www.math.kth.se/˜tkatchev
Contents
Chapter 1. Elliptic integrals and Jacobi’s theta functions 5
1.1. Elliptic integrals and the AGM: r eal case 5
1.2. Lemniscates and elastic curves 11
1.4. Theta functions: preliminaries 24
Chapter 2. General theory of doubly periodic functions 31
2.1. Preliminaries 31
2.2. Periods of analytic functions 33
2.3. Existence of doubly periodic functions 36
2.4. Liouville’s theorems 38
2.5. The Weierstrass function (z) 43
2.6. Modular forms 51
Bibliography 61
3
CHAPTER 1
Elliptic integrals and Jacobi’s theta functions
1.1. Elliptic integrals and the AGM: real case
1.1.1. Arclength of ellipses. Consider an ellipse with major and minor arcs 2a and
2b and eccentricity e := (a
2
b
2
)/a
2
[0, 1), e.g.,
x
2
a
2
+
y
2
b
2
= 1.
What is the arclength `(a; b) of the ellipse, as a function of a and b? There are two easy
(1) `(ra; rb) = r`(a; b), because rescaling by a factor r increases the arclength by the
same factor;
(2) `(a; a) = 2πa, because we know the circumference of a circle.
Of course, π is transcendental so it is d ebatable how well we understand it!
–1
–0.5
0
0.5
1
–2
–1 1
2
Figure 1. Ellipse x
2
+
y
2
4
= 1
The total arclength is four times the length of the piece in the ﬁrst quadrant, wher e we
have the relations
y = b
p
1 (x/a)
2
, y
0
(x) =
xb
a
2
1
p
1 (x/a)
2
.
5
Thus we obtain
`(a, b) = 4
Z
a
0
p
1 + y
02
(x) dx =
substituting z = x/a
= 4a
Z
1
0
r
1 ez
2
1 z
2
dx =
= 4a
Z
1
0
1 ez
2
p
(1 ez
2
)(1 z
2
)
dx.
This is an example of an elliptic integral of the second kind.
1.1.2. The simple pendulum. How do we compute the period of motion of a simple
pendulum? Suppose the length of the pendulum is L and the gravitational constant is g.
Let θ be the angle of the displacement of the pendulum from the vertical. The motion of
the pendulum is governed by a diﬀerential equation
θ
00
(t) =
g
L
sin θ(t).
In basic calculus and physics classes, this is traditionally linearized to
θ
00
(t) =
g
L
θ(t), θ 0,
so that the solutions take the form
θ(t) = A cos ωt + B sin ωt, ω =
r
g
L
.
We obtain simple harmonic motion with frequency ω and period 2π.
We shall consider the nonlinear equation, using a series of substitutions. First, note that
our equation integrates to
1
2
θ
02
ω
2
cos θ = const
Assume that the pendulum has a maximal displacement of angle θ = α; then θ
0
(α) = 0 so
we h ave
1
2
θ
02
= ω
2
(cos θ cos α),
and thus,
θ
0
= ±ω
p
2(cos θ cos α).
We take positive square root before the maximal displacement is achieved. Integrating again,
we ob tain
ωt =
Z
θ
0
p
2(cos φ cos α)
=
1
2
Z
θ
0
q
sin
2
α
2
sin
2
φ
2
.
Substituting
z =
sin
φ
2
sin
α
2
, ρ =
sin
θ
2
sin
α
2
, e = sin
2
α
2
[0, 1),
6
we ob tain
ωt =
Z
ρ
0
dz
p
(1 z
2
)(1 ez
2
)
.
At maximal displacement θ = α we have ρ = 1, so the ﬁrst time where maximal displacement
occurs is given by
T
4
=
1
ω
Z
1
0
dz
p
(1 z
2
)(1 ez
2
)
,
where T is the period of the oscillation (which is four times the time needed to achieve the
maximal displacement). These are examples of elliptic integrals of the ﬁrst kin d.
Finally, we should point out that actually computing the function θ(t) involves inverting
the function
ρ
Z
ρ
0
dz
p
(1 z
2
)(1 ez
2
)
.
1.1.3. The arithmetic-geometric mean iteration. The arithmetic-geometric mean
of two numbers a and b is deﬁned to be the common limit of the two sequences {a
n
}
n=0
and
{b
n
}
n=0
determined by the algorithm
a
0
= a, b
0
= b
a
n+1
=
a
n
+ b
n
2
, b
n+1
=
p
a
n
b
n
, n = 0, 1, 2 . . . ,
(1.1)
where b
n
+ 1 is always th e positive square root of a
n
b
n
.
Note that a
1
and b
1
are the respective arithmetic and geometric means of a and b, a
2
and
b
2
the corresponding means of a
1
and b
1
, etc. Thus the limit
M(a, b) := lim
n→∞
a
n
= lim
n→∞
b
n
(1.2)
really do es deserve to be called the arithmetic-geometric mean (AGM) of a and b. This
algorithm ﬁrst appeared in pap ers of Euler and Lagrange (sometime before 1785), bu t it
was Gauss who really discovered (in the 1790s at the age of 14) the amazing depth of this
subject. Unfortunately, Gauss published little on the AGM during his lifetime.
1
Theorem 1.1. Let a and b be positive real numbers. Then the limi ts in (1.2) do exist
and coincide.
Proof. We will assume that a b > 0, and we let {a
n
}
n=0
and {b
n
}
n=0
be as in (1.1).
The usual inequality between arithmetic and geometric means,
a
n
+ b
n
2
p
a
n
b
n
1
By May 30th, 1799, Gauss had observed, purely computationally, that
1
M(1,
2)
and
2
π
Z
1
0
dt
1 t
4
agreed to at least eleven (!) decimal places. He commented in his diary that this result ”will surely open
up a whole new ﬁeld of analysis” a claim vindicated by the subsequent directions of nineteenth-century
mathematics. The inverse of the above (indeﬁnite) integral is the lemniscate sine, a function Gauss studied
in some detail. He had recognized it as a doubly periodc function by the year 1800 and hence had anticipated
one of the most important developments of Abel and Jacobi: the inverse of algebraic integrals.
7
Figure 2. GAUSS Carl Friedrich (1777-1855)
immediately implies that a
n
b
n
for all n 0. Actually, much more is true: we have
a
1
a
2
. . . a
n
a
n+1
. . . b
n+1
b
n
. . . b
1
b
0
(1.3)
and
0 a
n
b
n
2
n
(a b). (1.4)
To prove (1.3), note that a
n
b
n
and a
n+1
b
n+1
imply
a
n
a
n
+ b
n
2
= a
n+1
b
n+1
=
p
a
n
b
n
b
n
,
and (1.3) follows. From b
n+1
b
n
we ob tain
a
n+1
b
n+1
a
n+1
b
n
= 2
1
(a
n
b
n
),
and (1.4) follows by induction. From (1.3) we see immediately that lim
n→∞
a
n
and lim
n→∞
b
n
exist, and (1.4) implies that the limits are equal.
Thus, we can use (1.2) to deﬁne the arithmetic-geometric mean M(a, b) of a and b.
Below we list the simple propert ies of the AGM.
Fact 1: M(a, a) = a;
Fact 2: M(a, b) = M(b, a);
Fact 3: M(a, 0) = 0;
Fact 4: M(a, b) = M(a
1
, b
1
) = M(a
2
, b
2
) = . . .;
Fact 5: M(λa, λ b) = λM(a, b);
Fact 6: M(a, b) = M
a+b
2
,
ab
.
In particular, the latter relation leads us to
M(1, x) = M
1 + x
2
,
x
,
which shows that the AGM f(x) := M(1, x) is a solution to the following functional equation
f(x) =
1 + x
2
f
2
x
1 + x
.
Our next result shows that the AGM is not as simple as indicated by what we have done
so far. We now get our ﬁrst glimpse of the depth of this subject.
8
Theorem 1.2 (Gauss, 1799). Let a and b are positive reals. Then
1
M(a, b)
=
2
π
Z
π/2
0
p
a
2
cos
2
φ + b
2
sin
2
φ
Proof 1. As before, we assume that a b > 0. Let I(a, b) denote the above integral,
and set µ = M(a, b). Thus we need to prove
I(a, b) =
π
2µ
.
The key st ep is to show that
I(a, b) = I(a
1
, b
1
). (1.5)
Let us introduce a new variable φ
0
such that
sin φ =
2a sin φ
0
a + b + (a b) sin
2
φ
0
. (1.6)
Note that 0 φ
0
π
2
corresponds to 0 φ
π
2
. To see this we consider the function
f(t) :=
2at
a + b + (a b)t
2
.
Then
f
0
(t) = 2a
(a + b (a b)t
2
)
(a + b + (a b)t
2
)
2
2ab
(a + b + (a b)t
2
)
2
> 0
which means that f(t) increasing in [0, 1]. On the other hand,
f(0) = 0, f(1) = 1,
which yields our claim.
Now, we note that
p
a
2
cos
2
φ + b
2
sin
2
φ
=
0
p
a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
. (1.7)
Indeed, one can ﬁnd from (1.6)
cos φ =
2 cos φ
0
p
a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
a + b + (a b) sin
2
φ
0
(1.8)
and it follows ( by straightforward manipulations) that
q
a
2
cos
2
φ + b
2
sin
2
φ = a
a + b (a b) sin
2
φ
0
a + b + (a b) sin
2
φ
0
. (1.9)
Then (1.7) follows from these formulas by taking the diﬀerential of (1.6).
Iterating (1.5) gives us
I(a, b) = I(a
1
, b
1
) = I(a
2
, b
2
) = . . . ,
so that
I(a, b) = lim
n→∞
I(a
n
, b
n
) = I(µ, µ) =
π
2µ
,
9
since the functions
1
p
a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
converge uniformly to the constant function
1
µ
.
Remark 1.1.1. Here we prove (1.7).
cos
2
φ = 1
4a
2
sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
=
(a + b)
2
+ 2(a
2
b
2
) sin
2
φ
0
+ (a b)
2
sin
4
φ
0
4a
2
sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
= (using our notation for a
1
and b
1
) =
=
4a
2
1
4(2a
2
1
b
2
1
) sin
2
φ
0
+ 4(a
2
1
b
2
1
) sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
=
4(a
2
1
cos
4
φ
0
+ 4b
2
1
sin
2
φ
0
cos
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
and (1.8) follows.
To prove (1.9) we note that
a
2
cos
2
φ + b
2
sin
2
φ =
4a
2
cos
2
φ
0
(a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
) + 4a
2
b
2
sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
= 4a
2
a
2
1
(1 sin
2
φ
0
)
2
+ b
2
1
sin
2
φ
0
(1 sin
2
φ
0
) + b
2
sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
= (using the old variables a and b) =
= a
2
(a + b)
2
(1 sin
2
φ
0
)
2
+ 4ab sin
2
φ
0
(1 sin
2
φ
0
) + 4b
2
sin
2
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
= a
2
(a + b)
2
2(a b)(a + b) sin
2
φ
0
+ (a b)
2
sin
4
φ
0
((a + b) + (a b) sin
2
φ
0
)
2
=
which implies (1.9).
Finally, (1.6) gives
cos φ = 2a
a + b (a b) sin
2
φ
0
(a + b + (a b) sin
2
φ
0
)
2
cos φ
0
0
.
We have for the left-hand side from (1.8)
cos φ = 2 cos φ
0
q
a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
a + b + (a b) sin
2
φ
0
which yields
q
a
2
1
cos
2
φ
0
+ b
2
1
sin
2
φ
0
= a
a + b (a b) sin
2
φ
0
a + b + (a b) sin
2
φ
0
0
=
=
q
a
2
cos
2
φ + b
2
sin
2
φ
0
and (1.7) is proven.
10
Remark 1.1.2. Another proof is d ue to Carlson . It uses the representation
2
π
Z
π/2
0
p
a
2
cos
2
φ + b
2
sin
2
φ
=
1
π
Z
+
−∞
dt
p
(a
2
+ t
2
)(b
2
+ t
2
)
(1.10)
with the further substitution
u :=
1
2
t
ab
t
.
Exercise 1.1.1. Consider the harmonic-geometric mean iteration
α
n+1
=
2α
n
β
n
α
n
+ β
n
, β
n+1
=
p
α
n
β
n
.
Show, for α
0
, β
0
(0, ), that the above iteration converges to
H(α
0
, β
0
) =
1
M(1
0
, 1
0
)
.
Exercise 1.1.2. Prove (1.10) and the recurrence relation
Z
+
−∞
dt
p
(a
2
+ t
2
)(b
2
+ t
2
)
=
Z
+
−∞
dt
p
(a
2
1
+ t
2
)(b
2
1
+ t
2
)
.
1.2. Lemniscates and elastic curves
. . . Today, the elastic curve has been largely forgot-
ten, and he lemniscate has suﬀered the worse fate
of being r elegated to the polar coordinates section
of calculus books. There it sits next to the formula
for arc length in polar coordinates, which can never
be applied to the lemniscate since such texts know
nothing of elliptic integrals. . .
D.A. Cox, .
1.2.1. Arclength of lemniscate. A lemniscate
2
was discovered by Jacob Bernoulli
in 1694.
Figure 3. The lemniscate
He gives th e equation in the form
(x
2
+ y
2
)
2
= 2a
2
(x
2
y
2
) (1.11)
2
Animation and formulas: http://www.mathcurve.com/courbes2d/lemniscate/lemniscate.shtml
Formulas and more calculations: http://mathworld.wolfram.com/Lemniscate.html
11
and explains that the curve has ”the form of a ﬁgure 8 on its side, as of a band folded into
a knot, or of a lemniscus, or of a knot of a French ribbon”
3
.
Figure 4. Jacob Bernoulli (1654 1705)
A polar curve also called Lemniscate of Bernoulli which is the locus of points the product
of whose distances from two points (called the foci) is a constant. Letting the foci be located
at (±a, 0), the Cartesian equation is
[(x a)
2
+ y
2
][(x + a)
2
+ y
2
] = a
4
.
The polar coordinates are given by
r
2
= 2a
2
cos(2θ). (1.12)
Let now ﬁx a = 1/
2 so that (1.12) can be written as r
2
(θ) = cos 2θ. Using th e formula
for arc length in polar coordinates, see that the total arc length L is
L = 4
Z
π/4
0
(r
2
+ r
02
)
1/2
= 4
Z
π/4
0
cos 2θ
.
The substitution cos 2θ = cos
2
φ transforms this to the integral
L = 4
Z
π/2
0
p
1 + cos
2
φ
= 4
Z
π/2
0
p
2 cos
2
φ + sin
2
φ
=
2π
M(
2, 1)
,
which links the Gauss AGM M(
2, 1) and the arc length of the lemniscate.
Finally, letting t = cos φ we obtain
L = 4
Z
π/2
0
p
2 cos
2
φ + sin
2
φ
= 4
Z
1
0
dt
1 t
4
. (1.13)
1.2.2. Elastic Curves. More interesting is that the integral in the right h and side of
(1.13) had been discovered by Jacob Bernoulli three years earlier in 1691.
This was when Bernoulli worked out the equation of the so-called elastic curve. The
situation is as follows: a thin elastic rod is bent until the two ends are perpendicular to a
given line H. After introducing cartesian coordinates as indicated on Figure 5 and letting
3
In 1694 Jacob Bernoulli published a curve in Acta Eruditorum. Following the protocol of his day, he
gave this curve the Latin name of lemniscus, which translates as a pendant ribbon to be fastened to a victor’s
garland. He was unaware that his curve was a special case of the Ovals of Cassini. His inves tigations on the
length of the arc laid the foundation for later work on elliptic functions.
12
A
H
Rod
–1
–0.5
0
0.5
1
y
0.5
1 1.5
2
x
Figure 5. Elastic curve
a denote 0A, Bernoulli was able to show that the upper half of the curve is given by the
equation
y =
Z
x
0
t
2
dt
a
4
t
4
, (1.14)
where 0 x a.
It is convenient t o assume that a = 1. But as soon as this is done, we no longer know
how long the rod is. In fact, (1.14) implies that the arc length from the origin to a point
(x, y) on the rescaled elastic curve is
`(x) =
Z
x
0
(1 t
4
)
1/2
dt. (1.15)
Thus t he half-length of the whole rod is
` :=
Z
1
0
dt
1 t
4
.
How did Bernoulli get from here to the lemniscate? He was well aware of the transcen-
dental natu re of the elastic curve, and so he used a standard seventeenth century trick to
make things m ore manageable: he sought an algebraic curve whose rectiﬁcation should agree
with the rectiﬁcation of the elastic curve.
Since Bernoulli’s solution involved the arc length of the elastic curve, it was natural for
him to seek an algebraic curve with the same arc length. Very shortly thereafter, he found
the equation of the lemniscate. So the arc length of the lemniscate was known well before
the curve itself.
1.2.3. Euler’s identity. Throughout the 18th century the elastic curves and the lem-
niscate app eared in many papers. A lot of work was done on the integrals (1.14) and (1.15).
A notable work on the elastic curve was Euler’s paper of 1786. Namely, Euler gives ap-
proximations to the above integrals and, more importantly, proves the following amazing
result
13
Theorem 1.3 (Euler’s Identity).
Z
1
0
dt
1 t
4
·
Z
1
0
t
2
dt
1 t
4
=
π
4
. (1.16)
The proof is given in Exercise 1.2.2. We prove a more general assertion following 
Theorem 1.4 (The generalized elastic curves, ). Let
f
n
(x) :=
Z
x
0
t
n
dt
1 t
2n
be the generalized elastic curve. Let us denote by R
n
= f
n
(1) the so called main radius, and
by L
n
the length of the curve from x = 0 to x = 1. Then
L
n
R
n
=
π
2n
. (1.17)
Remark 1.2.1. One can easily observe that for n = 1 (1.17) is the well-known identity
since
L
1
=
π
2
, R
1
= 1.
Proof. We have
R
n
=
Z
1
0
t
n
dt
1 t
2n
,
and one can easily ﬁnd that
L
n
=
Z
1
0
dt
1 t
2n
.
Integrate the relation
d(t
k
1 t
2n
) =
kt
k1
t
k1
(k + n)t
2n+k1
1 t
2n
dt
from 0 to 1 to produce the recursive formula
Z
1
0
t
k1
dt
1 t
2n
=
k + n
k
Z
1
0
t
2n+k1
dt
1 t
2n
. (1.18)
The value k = n + 1 in (1.18) yields
R
n
=
2n + 1
n + 1
Z
1
0
t
3n
dt
1 t
2n
.
Then the value k = 3n + 1 produces
R
n
=
4n + 1
3n + 1
Z
1
0
t
5n
dt
1 t
2n
,
so we have
R
n
=
2n + 1
n + 1
×
4n + 1
3n + 1
Z
1
0
t
5n
dt
1 t
2n
.
Iterating we obtain, after m steps,
R
n
=
m
Y
j=1
2jn + 1
(2j 1)n + 1
×
Z
1
0
t
(2m+1)n
dt
1 t
2n
. (1.19)
14
The next step is to justify t he passage to the limit in (1.19) as m , with n ﬁxed. Observe
that the left hand side is independent of m, so it remains R
n
after m . The diﬃculty
in passing to the limit is that the product in (1.19) diverges. The general term p
j
satisﬁes
1 p
j
=
n
(2j 1)n + 1
and the divergence of the product follows from that of the harmonic series. The divergence
is cured by introducing scaling factors both in the integral and the product.
Proposition 1.1. The functions
1
2m + 1
m
Y
j=1
2jn + 1
(2j 1)n + 1
and
2m + 1
Z
1
0
t
(2m+1)n
dt
1 t
2n
have non-zero limits as m .
Therefore from (1.19) we obtain
R
n
= lim
m→∞
2m
Y
j=1
(jn + 1)
(1)
j
×
Z
1
0
t
(2m+1)n
dt
1 t
2n
,
where we have employed
2m
Y
j=1
(jn + 1)
(1)
j
=
m
Y
j=1
2jn + 1
(2j 1)n + 1
in order to simplify the notation. A similar argument shows that
L
n
=
m
Y
j=1
(2j 1)n + 1
2(j 1)n + 1
×
Z
1
0
t
2mn
dt
1 t
2n
= lim
m→∞
2m
Y
j=1
(jn + 1)
(1)
j+1
×
Z
1
0
t
2mn
dt
1 t
2n
The ﬁnal step is to introduce the auxiliary quantities
A
n
:=
Z
1
0
t
n1
dt
1 t
2n
and B
n
:=
Z
1
0
t
2n1
dt
1 t
2n
.
We now show that the quotient L
n
/A
n
can be evaluated explicitly and that the value of A
n
is elementary. This produces an expression for L
n
. A similar statement holds for R
n
/B
n
and
B
n
.
Observe ﬁ rst that (after change of the variable)
A
n
=
Z
1
0
t
n1
dt
1 t
2n
=
1
n
Z
1
0
dt
1 t
2
=
π
2n
, (1.20)
and similarly B
n
= 1/n. Now consider the recursion (1.18) for odd multiples of n to
A
n
= lim
m→∞
2m
Y
j=1
(jn)
(1)
j
×
Z
1
0
t
(2m+1)n1
dt
1 t
2n
15
and similarly the even multiples of n yield
B
n
=
1
n
lim
m→∞
2m+1
Y
j=1
(jn)
(1)
j+1
×
Z
1
0
t
(2m+1)n1
dt
1 t
2n
in the exact manner as the derivation of (1.19). Therefore using the last identities, and
passing to the limit as m so that the integrals disappear, we obtain
L
n
A
n
=
Y
j=1
h
(jn + 1)
(1)
j+1
× (jn)
(1)
j+1
i
so (1.20) yields
L
n
=
π
2n
×
Y
j=1
h
(jn + 1)
(1)
j+1
× (jn)
(1)
j+1
i
.
Similarly, using B
n
= 1/n,
R
n
= ×
Y
j=1
h
(jn + 1)
(1)
j
× (jn)
(1)
j
i
.
The formula Rn × Ln = π/2n follows directly from here.
Exercise 1.2.1. Prove Proposition 1.1.
Exercise 1.2.2. Give another proof of Theorem 1.4 by using the B and Γ Euler’s func-
tions:
B(α, β) =
1
Z
0
(1 t)
α1
t
β1
dt =
Γ(α)Γ(β)
Γ(α + β)
, (1.21)
and the fact that Γ(1/2) =
π.
1.2.4. Addendum: The lemniscate and its ”relatives”.
1.2.4.1. Circ le and lemniscate. There are several methods for drawing a lemniscate. The
easiest is illustrated below. Draw a circle and then extend a diameter to become a secant. The
center of the lemniscate O will be
2 times the radius of the circle. Through O draw several
segments cutting the circle. The pattern of the lemniscate emerges in the ﬁrst quadrant (see
Figure 6).
Exercise 1.2.3. Prove the mentioned property for the unit circle. Hint: use the secant
line equation (with ﬁxed angle α)
(x, y) = (
2 + t cos α, t sin α), α
h
π
4
,
π
4
i
,
where the interval t [τ
1
(α), τ
2
(α)] is deﬁned by substitution in the circle equation x
2
+y
2
=
1. Then the required equation of the lemniscate is given by
(X, Y ) = (τ
1
(α) τ
2
(α)) × (cos α, sin α).
It’s also worth noting that the lemniscate is the inverse (in the sense of inversive geometry)
of the hyperbola relative to the circle of radius k = a
2 where a is deﬁned by (1.11). In
other words, if we draw a line emanating from the origin and it strikes the lemniscate at the
radius s, then it strikes the hyperbola at the radius R where sR = k
2
.
16
Figure 6. The lemniscate and the circle
1.2.4.2. A lemniscate ”machine”. Another method is based on the mechanical interpre-
tation of the main lemniscate property and is illustrated by Figure 7.
Figure 7. lemniscate ”machine” 
Exercise 1.2.4. Give an ”explanation” of the lemniscate machine.
1.2.4.3. Cassinian Ovals. Jacob Bernoulli was not aware that the curve he was describing
was a special case of Cassini Ovals which had been described by Cassini in 1680.
Cassinian oval describe a family of curves. It is deﬁned as the locus of points P such
that the product of distances |P F
1
||P F
2
| = b
2
is constant. Here F
1
and F
2
are two ﬁxed
points (foci) and b is a constant. It is analogous to the deﬁnition of ellipse, where sum of
17
two distances is replace by product. Let the distance between the foci be 2a. Then a special
case is the lemniscate of Bernoulli when a = b.
Exercise 1.2.5. Prove that the polar representation of the Cassinian Ovals is given by
r
4
+ a
4
2r
2
a
2
cos 2θ = b
4
.
Cassinian ovals are the intersection of a torus and a plane in certain position. Let a be
the inner radius of a torus whose generating circle has radius R (see Figure 8). Cassinian
oval is the intersection of a plane parallel to the torus’ axis and R distant from it. If a = 2R,
then it is the lemniscate of Bernoulli. Note that these tori in the ﬁgure are not identical.
(Obs!: Arbitrary slice of a torus are not Cassinian ovals).
Figure 8. Cassinian ovals as intersection of a torus and a plane
Exercise 1.2.6. Prove th e preceding assertion. Hint: Use the Cartesian representation
of the torus
(
p
x
2
+ y
2
a)
2
+ z
2
= R
2
.
1.3.1. Fagnano’s Theorem on the lemniscate. Unlike the elastic curve, the story of
the lemniscate in the 18th century is well known, primarily because of the key role it played
in the development of the theory of elliptic integrals. One early worker was Giulio Carlo
Fagnano (1682–1766). He, following some ideas of Johann Bernoulli, Jacob’s younger
brother, studied the ways in which arcs of ellipses and hyperbolas can be related.
One result, known as Fagnano’s Theorem, states that the sum of two appropriately
chosen arcs of an ellipse can be computed algebraically in terms of the coordinates of the
points involved
4
. He also worked on the lemniscate, starting with the problem of halving
4
These researches of Fagnano’s were published in the period 1714–1720 in an obscure Venetian journal
and were not widely known. In 1750 he had his work republished, and he sent a copy to the Berlin Academy.
It was given to Euler for review on December 23, 1751. Less than ﬁve weeks later, on January 27, 1752,
Euler read a paper giving new derivations for Fagnano’s results on elliptic and hyperbolic arcs. By 1753
elliptic integrals.
18
that portion of the arc length of the lemniscate which lies in one quadrant. Subsequently he
found methods for dividing this arc length into n equal pieces, where n = 2
m
, 3·2
m
or 5 ·2
m
.
We formulate the simplest case of Fagnano’s Theorem the duplication of the lemniscate
arc length.
Theorem 1.5 (Fagnano’s Doubling Theorem). Let 0 < u <
p
2 1 and
r =
2u
1 u
4
1 + u
4
.
Then
Z
r
0
dt
1 t
4
= 2
Z
u
0
dt
1 t
4
.
Proof. First we note that the function
r = f(u) :=
2u
1 u
4
1 + u
4
have as its derivative
f
0
(u) = 2
(u
4
+ 2 u
2
1) (u
4
2 u
2
1)
(1 + u
4
)
2
1 u
4
,
so it increasing in [0, u
0
] with u
0
being the least positive root of f
0
(u) = 0. Clearly, u
0
=
p
2 1.
On the other hand
1 f(u)
4
= 1
16u
4
(1 u
4
)
2
(1 + u
4
)
4
=
(u
4
+ 2 u
2
1)
2
(u
4
2 u
2
1)
2
(1 + u
4
)
4
.
It follows t hat
df
p
1 f
4
(u)
= 2
du
1 u
4
and the result follows.
1.3.2. Addition theorems. The simplest example of a function which has an algebraic
addition theorem is the exponential function
φ(u) = e
u
.
It follows t hat
e
u
· e
v
= e
u+v
,
or
φ(u) · φ(v) = φ(u + v).
Such an equation oﬀers a means of determining the value of the function for the sum of
two quantities as arguments, when the values of the function for the two arguments taken
singly are known.
It is called an addition theorem.
In the example just cited the relation among φ(u), φ(v) and φ(u+v) is expressed through
an algebraic equation, and consequently the addition theorem is called algebraic addition
theorem.
19
Figure 9. Leonard EULER (1707-1783)
The sine function has the algebraic addition theorem
sin(u + v) = sin u cos v + cos u sin v =
= sin u
p
1 sin
2
v + sin v
p
1 sin
2
u.
(1.22)
We also have
tan(u + v) =
tan u + tan v
1 tan u tan v
Another result is the previous Fagnano’s duplication theorem, which can be reformulated
as follows: let φ(u) be deﬁned as the solution to
u =
Z
φ(u)
0
dt
1 t
4
.
Then
φ(2u) = φ(u + u) =
2φ(u)
p
1 φ(u)
4
1 + φ(u)
4
.
Theorem 1.6 (Euler’s Addition Theorem). Let
f(x) := (1 x
2
)(1 k
2
x
2
).
Then
Z
x
0
dt
p
f(t)
+
Z
y
0
dt
p
f(t)
=
Z
z
0
dt
p
f(t)
, (1.23)
where
z =
x
p
f(y) + y
p
f(x)
1 k
2
x
2
y
2
. (1.24)
Proof. We follow a method of proving the Euler theorem due to Darboux [1, p. 73].
Let us consider the equation
dx
p
(1 x
2
)(1 k
2
x
2
)
+
dy
p
(1 y
2
)(1 k
2
y
2
)
= 0. (1.25)
20
Obviously, that (1.25) deﬁnes a level set of the function z = z(x, y) deﬁned by (1.23). If we
set
u =
Z
x
0
dt
p
f(t)
,
v =
Z
y
0
dt
p
f(t)
,
then the integral identity (1.23) can be represented in the form
u + v = A,
where A is a suitable constant.
On the other hand, equation (1.25) can be replaced by the system
dx
dt
=
p
(1 x
2
)(1 k
2
x
2
)
dy
dt
=
p
(1 y
2
)(1 k
2
y
2
).
(1.26)
Squaring (1.26), we get
(
dx
dt
2
= (1 x
2
)(1 k
2
x
2
)
dy
dt
2
= (1 y
2
)(1 k
2
y
2
).
(1.27)
Let us now diﬀerentiate these equations:
d
2
x
dt
2
= x(2k
2
x
2
1 k
2
),
d
2
y
dt
2
= y(2k
2
y
2
1 k
2
).
This implies
y
d
2
x
dt
2
x
d
2
y
dt
2
= 2k
2
xy(x
2
y
2
),
or
d
dt
y
dx
dt
x
dy
dt
= 2k
2
xy(x
2
y
2
). (1.28)
On the other hand, it follows from (1.27) that
y
2
dx
dt
2
x
2
dy
dt
2
= (y
2
x
2
)(1 k
2
x
2
y
2
). (1.29)
Dividing (1.28) by (1.29), we get
d
dt
y
dx
dt
x
dy
dt
y
dx
dt
x
dy
dt
=
2k
2
xy
y
dx
dt
+ x
dy
dt
k
2
x
2
y
2
1
,
or
d
dt
ln
y
dx
dt
x
dy
dt
=
d
dt
ln(k
2
x
2
y
2
1).
Thus, we have
y
dx
dt
x
dy
dt
= C(k
2
x
2
y
2
1).
21
Taking (1.26) into account, we get
x
p
(1 y
2
)(1 k
2
y
2
) + y
p
(1 x
2
)(1 k
2
x
2
)
1 k
2
x
2
y
2
= C.
This is the desired algebraic form of the integral of the equation (1.25). The theorem is
proved.
Corollary 1.1. Let k = 0. Then the assertion of the theorem is equivalent to (1.22).
1.3.3. J acobi’s functions: preliminaries. The last considerations lead us to the most
popular Jacobian elliptic functions which are
sine amplitude elliptic function sn(x, k),
cosine amplitude elliptic function cn(x, k),
delta amplitude elliptic function dn(x, k).
These functions may be deﬁned via the inverse of the incomplete elliptic integrals as
follows:
x =
sn(x,k)
Z
0
dt
p
(1 t
2
)(1 k
2
t
2
)
x =
cn(x,k)
Z
1
dt
p
(1 t
2
)(k
02
+ k
2
t
2
)
x =
dn(x,k)
Z
1
dt
p
(1 t
2
)(t
2
k
02
)
.
The second argument of the functions k is a modulus of the elliptic function and
k
0
:=
1 k
2
is a complimentary modulus. The eight remaining Jacobian elliptic functions can be conve-
niently deﬁned via th e general identity relations
fg(x, k) =
fe(k, x)
ge(k, x)
e,f,g=s,c,d,n,
where ﬀ(x, k) is interpreted as unity.
In this section we examine the simplest property only. To treat Jacobi’s function in more
detail we need to extend them into complex plane which will be given in the further sections.
Proposition 1.2. The following identities hold
sn
2
(x, k) + cn
2
(x, k) = 1, (1.30)
dn
2
(x, k) + k
2
sn
2
(x, k) = 1. (1.31)
22
Proof. Let 0 x 1 be ﬁxed, and u := sn(x, k), v := cn(x, k). Then we have by the
deﬁnition
x =
v
Z
1
dt
p
(1 t
2
)(k
02
+ k
2
t
2
)
=
s =
1 t
2
dt =
sds
1s
2
=
=
1v
2
Z
0
ds
p
(1 s
2
)(1 k
2
s
2
)
,
which clearly implies u =
1 v
2
. Thus, (1.30) is proven. The second identity is proved in
a similar way.
Exercise 1.3.1. Show that
sn(x, 0) = sin x, sn(x, 1) = tanh x,
cn(x, 0) = cos x, cn(x, 1) =
1
cosh x
,
dn(x, 0) = 1, dn(x, 1) =
1
cosh x
.
Now, we have an important consequence of Theorem 1.6
Corollary 1.2. Addition/substraction formulae for sine, cosine and delta amplitude
Jacobian functions are
sn(x ± y; k) =
sn(x; k) cn(y; k) dn(y; k) ± cn(x; k) dn(x; k) sn(y; k)
1 k
2
sn
2
(x; k) sn
2
(y; k)
cn(x ± y; k) =
cn(x; k) cn(y; k) ± sn (x; k) d n(x; k) sn (y; k) dn(y; k)
1 k
2
sn
2
(x; k) sn
2
(y; k)
dn(x ± y; k) =
dn(x; k) dn(y; k) k
2
sn(x; k) cn(x; k) sn(y; k) cn(y; k)
1 k
2
sn
2
(x; k) sn
2
(y; k)
The following p roperty provides an easy application of the Addition Theorem.
Proposition 1.3. Let
K = K(k) :=
Z
1
0
dz
p
(1 x
2
)(1 k
2
x
2
)
be the complete integral of the ﬁrst kind. Then the following identities hold
sn(K, k) = 1, sn(
K
2
, k) =
1
1 +
k
2
1
. (1.32)
Clearly, that K(k) plays the role of
π
2
for the Jacobi sine function. Hint for the pr oof:
write
sn(
K
2
, k) = sn(K
K
2
, k).
23
1.4. Theta functions: preliminaries
1.4.1. Theta functions as solutions of the Heat Conduction Problem. Theta
functions appear appear in Bernoulli’s Ars Conjectandi  and in the number-theoretic
investigations of Euler  and Gauss , but come into full ﬂower only in Jacobi’s
Fundamenta Nova .
We shall introduce the theta functions by considering a speciﬁc heat conduction problem.
Namely, in this way the theta functions occur in J. Fourier’s La Th´eorie Analytique de la
Chaleur .
Let θ be the temperature at the time t at any point in a solid material whose conduction
properties are uniform and isotropic. Then, if ρ is the material’s density, s is its speciﬁc
heat, and k its thermal conductivity, θ satisﬁes the partial diﬀerential equation
κθ =
θ
t
, (1.33)
where κ = s/ρ is termed the diﬀusivity and is the Laplace operator. In the special case
where there is no variation of temperature in the x- and y-directions of a rectangular Carte-
sian frame Oxyz, the heat ﬂow is everywhere parallel to the z-axis and the heat conduction
reduces to the form
κ
2
θ
z
2
=
θ
t
, (1.34)
and θ = θ(z, t).
The speciﬁc problem we shall study (in an ideal form) is the ﬂow of heat in an inﬁ nite slab
of material, bounded by the planes z = 0, z = π, when the conditions over each boundary
plane are kept uniform at every time t. The heat ﬂow is then entirely in the z-direction and
equation (1.34) is applicable.
First, suppose the boundary conditions are that the faces of the slab are maintained at
zero temperature, i.e. θ = 0 for z = 0, π and all t. Initially, at t = 0 suppose
θ(z, 0) = f(z), 0 < z < π.
Then the method of separation of variables leads to the solution
θ(z, t) =
X
n=1
b
n
e
n
2
κt
sin nz, (1.35)
where b
n
are Fourier coeﬃcients determined by the equation
b
n
=
2
π
Z
π
0
f(z) sin nz dz. (1.36)
Exercise 1.4.1. Show the validity of (1.35). Hint: consider the Fourier expansion of
f(z) in the sin-series and show that e
m
2
κt
sin mz solves (1.34) with f
m
(z) = sin mz.
In the special case where
f(z) = πδ(z
1
2
π)
(with δ(z) to be the Dirac’s unit impulse function), the slab is initially at zero temperature
everywhere, except in the neighborhood of the midplane z =
π
2
, where the temperature is
24
very high. To achieve this high temperature, it will be necessary to inject a quantity of heat
h (joules) per unit area into this plane to raise its temperature from zero h is given by
h = ρsπ
Z
π/2+0
π/20
δ(z
1
2
π) dz = ρsπ. (1.37)
We now calculate that
b
n
= 2
Z
π
0
δ(z
1
2
π) sin nz dz = 2 sin
2
.
Thus, heat diﬀusion over the slab is governed by the equation
θ(z, t) = 2
X
n=0
(1)
n
e
(2n+1)
2
κt
sin(2n + 1)z. (1.38)
Writing
q := e
4κt
the solution (1.38) assumes the form
θ = θ
1
(z, q) = 2
X
n=0
(1)
n
q
(n+1/2)
2
sin(2n + 1)z. (1.39)
Deﬁnition 1.4.1. The function θ
1
(z, q) given by (1.39) is the ﬁrst theta function of
Jacobi.
1.4.2. Convergence property. The main technical result is as follows
Proposition 1.4. The ﬁrst theta function θ
1
(z, q) is deﬁned by the series (1.39) for all
complex values z and q such that |q| < 1. Moreover, the series converges uniformly in any
strip Y Im z Y , where Y > 0.
Proof. Replacing the sine function by its Euler representation by exponentials we obtain
the Nth partial sum of the series in (1.39)
S
N
:= 2
N
X
n=0
(1)
n
q
(n+1/2)
2
sin(2n + 1)z
=
q
1/4
i
N
X
n=0
(1)
n
q
n
2
+n
(e
(2n+1)zi
e
(2n+1)zi
) =
=
q
1/4
i
N
X
n=N
(1)
n
q
n
2
+n
e
(2n+1)zi
.
To establish convergence, let u
n
denote the nth term of the latter series. Then for n > 0 we
have
|u
n+1
|
|u
n
|
= |q
2n+2
e
2zi
| = |q|
2n+2
e
2y
, where z = x + iy. (1.40)
As n +, since |q| < 1, this ratio tends to zero and, by D’Alembert’s test, therefore, the
series converges at +. The similar argument shows that the series converges at −∞ and
the assertion follows.
25
Now, let us suppose that z is in the strip Y Im z Y , Y > 0. Then yields again, by
D’Alembert’s test and majorant principle that S
N
converges uniformly.
Corollary 1.3. θ
1
(z, q) is an integral holomorphic function of z.
Corollary 1.4. θ
1
(z, q) is a 2π-peri odic function of z.
Remark 1.4.1. An alternative notation (Gauss’ form) is to write
q = e
τ
, (1.41)
where now th e imaginary part of τ must be positive to give |q| < 1:
Re τ > 0. (1.42)
In this notation we have
1
(z|q) =
X
n=−∞
(1)
n
e
(n + 1/2)
2
π + (2n + 1)iz
.
1.4.3. Four Theta Functions. Incrementing z by a quarter period, we deﬁne the
second theta function θ
2
thus:
θ
2
(z, q) := θ
1
(z +
π
2
, q) =
= 2
+
X
n=0
q
(n+1/2)
2
cos(2n + 1)z
=
+
X
n=−∞
q
(n+1/2)
2
e
(2n+1)zi
.
Evidently, that θ
2
is an even integral function of z with period 2π. The next pair is
θ
3
(z, q) =
+
X
n=−∞
q
n
2
e
2inz
, (1.43)
and
θ
4
(z, q) = θ
3
(z
π
2
, q) =
+
X
n=−∞
(1)
n
q
n
2
e
2inz
.
The latter two functions are periodic (of z) with period π which follows from their Fourier
expansions
θ
3
(z, q) = 1 + 2
X
n=1
q
n
2
cos 2nz,
θ
4
(z, q) = 1 + 2
X
n=1
(1)
n
q
n
2
cos 2nz.
(1.44)
26
1.4.4. Theta functions and the AGM. Now we return to the central Gauss’ obser-
vation in his AGM treatises: the AGM solution in terms of theta functions. To do this we
restrict ourselves by the reduced theta series. Namely, we suppose that
z = 0,
so, our previous formulae p rovide the following deﬁnition.
The basic functions are deﬁned for |q| < 1 by
P (τ) := θ
2
(q) =
X
n=−∞
q
(n+1/2)
2
= 2q
1/4
+ 2q
9/4
+ 2q
25/4
+ . . . ,
Q(τ) := θ
3
(q) =
X
n=−∞
q
n
2
= 1 + 2q
1
+ 2q
4
+ 2q
9
+ 2q
16
+ . . . ,
R(τ) := θ
4
(q) =
X
n=−∞
(1)
n
q
n
2
= 1 2q
1
+ 2q
4
2q
9
+ 2q
16
. . . ,
with θ
j
(0) = 0 and τ as in (1.41).
Theorem 1.7 (The AGM representation via theta functions).
θ
2
3
(q) + θ
2
4
(q)
2
= θ
2
3
(q
2
),
q
θ
2
3
(q)θ
2
4
(q) = θ
2
4
(q
2
).
(1.45)
Proof. First we observe that
θ
4
(q) = θ
3
(q),
which yields
θ
3
(q) + θ
4
(q) = 2
X
n even
q
n
2
= 2θ
3
(q
4
). (1.46)
Also
θ
2
3
(q) =
X
n=−∞
X
m=−∞
q
n
2
+m
2
=
X
n=0
r
2
(n)q
n
, (1.47)
where r
2
(n) counts the number of ways of writing
n = j
2
+ k
2
. (1.48)
Here we distinguish sign and permutation [so that, for example, r
2
(5) = 8 since 5 = (±2)
2
+
(±1)
2
= (±1)
2
+ (±2)
2
] and set r
2
(0) := 1. Similarly we have
θ
2
4
(q) =
X
n=−∞
X
m=−∞
(1)
n+m
q
n
2
+m
2
=
X
n=0
(1)
n
r
2
(n)q
n
, (1.49)
since n
2
+ m
2
n + m mod 2.
Now we claim that
r
2
(2n) = r
2
(n). (1.50)
To prove this identity we ﬁx a number n 1 (the case n = 0 is trivial) and note that
2(a
2
+ b
2
) = (a b)
2
+ (a + b)
2
.
27
Then evidently a pair (a, b) solves (1.48) for n if and only if (A, B) := (a b, a + b) d oes
(1.48) for 2n. Clearly, that the correspondence
(a, b) (A, B) = (a, b)
1 1
1 1
is bijective which proves our claim.
It follows from ( 1.47) that
θ
2
3
(q) + θ
2
4
(q) = 2
X
n=0
r
2
(2n)q
2n
= 2θ
2
3
(q
2
). (1.51)
Also, (1.46) and (1.51) allow u s to solve for θ
3
(q)θ
4
(q):
θ
3
(q)θ
4
(q) =
1
2
(θ
3
(q) + θ
4
(q))
2
1
2
(θ
2
3
(q) + θ
2
4
(q)) =
= 2θ
2
3
(q
4
) θ
2
3
(q
2
) =
= again (1.51) =
= θ
2
4
(q
2
).
The theorem follows.
The main identity (1.45) bears an obvious resemblance with the AGM. Namely, we have
Corollary 1.5. Let |q| < 1. Then
M(θ
2
3
(q), θ
2
4
(q)) = M(θ
2
3
(q
2
), θ
2
4
(q
2
)) = . . . = M(θ
2
3
(q
2
n
), θ
2
4
(q
2
n
)) = . . .
(1.52)
Since θ
3
(0) = θ
3
(0) = 1 we easily arrive at
Corollary 1.6. Let |q| < 1. Then
M(θ
2
3
(q), θ
2
4
(q)) = 1.
(1.53)
Another important property is
Corollary 1.7 (Jacobi’s identity).
θ
4
3
(q) = θ
4
4
(q) + θ
4
2
(q).
(1.54)
Proof. We have
θ
2
3
(q) θ
2
3
(q
2
) =
X
n=0
r
2
(n)q
n
X
n=0
r
2
(2n)q
2n
=
= by (1.50) =
X
n=0
r
2
(2n + 1)q
2n+1
.
28
On the other hand,
X
n=0
r
2
(2n + 1)q
2n+1
= 2
X
k,m=−∞
k+modd
q
m
2
+k
2
=
= (setting k = i j, m = i + j + 1) =
=
X
i,j=−∞
(q
2
)
(i+1/2)
2
+(j+1/2)
2
= θ
2
2
(q
2
).
Hence
θ
2
3
(q
2
) + θ
2
2
(q
2
) = θ
2
3
(q),
which with t he ﬁrst identity in (1.45) produce
θ
2
3
(q
2
) θ
2
2
(q
2
) = θ
2
4
(q).
Now (1.53) follows from the last two identities and the second identity in (1.45).
Let us deﬁne
k := k(q) :=
θ
2
2
(q)
θ
2
3
(q)
.
Then (1.53) shows th at
k
0
:=
1 k
2
=
θ
2
4
(q)
θ
2
3
(q)
.
Theorem 1.8 (The AGM representation via theta functions, II). Let 0 < k <
1 is given. The AGM satisﬁes
M(1, k
0
) = θ
2
3
(q), for k
0
=
θ
2
4
(q)
θ
2
3
(q)
,
(1.55)
where q is the unique solution in (0, 1) to k = θ
2
2
(q)
2
3
(q) and k
2
+ k
02
= 1.
29
CHAPTER 2
General theory of doubly periodic functions
2.1. Preliminaries
2.1.1. Holomorphic functions. Here we summarize the well-known facts about the
analytic functions we need in the sequel.
We distinguish the ﬁnite complex plane C and its compactiﬁcation C = C {∞}, i.e.
the Riemann sphere. We use z = x + iy to indicate a complex number which plays the role
of a complex variable in what follows. As usually we set Re z = x and Im z = y for the real
and imaginary parts of z respectively.
The conjugate to z number is denoted by ¯z = x iy, and the modulus is deﬁned as the
following positive square root |z| =
z¯z. The argument of z 6= 0 is a multivalued function
arg z which is deﬁned by
z = |z|e
i arg z
.
A function f(z) of one complex variable z is called an analytic (or holomorphic) function
in an open set D C if it admits the following expansion in converging power series
f(z) = a
0
+ a
1
(z z
0
) + a
2
(z z
0
)
2
+ . . . ,
where the disk |z z
0
| < r is contained in D. In this case we have for the radius of
convergence:
0 < r R(z
0
) := [lim sup
n→∞
|a
n
|
1
n
]
1
.
The Taylor coeﬃcients a
n
are found by
a
n
=
f
(n)
(z
0
)
n!
,
and thus
f(z) =
X
n=0
f
(n)
(z
0
)
n!
(z z
0
)
n
, |z z
0
| < R(z
0
).
A function f(z) is said to be analytic at z
0
is it is analytic a small neighborhood of z
0
.
The point z
0
D is called a zero of f(z) of order N 1 if
f(z) = a
N
(z z
0
)
N
+ a
N+1
(z z
0
)
N+1
+ a
N+2
(z z
0
)
N+2
+ . . . , a
N
6= 0.
An equivalent condition is that f(z) = (z z
0
)
N
g(z) where g(z) is an analytic function at
z
0
and g(z
0
) 6= 0.
Cauchy Integral Theorem. Let f(z) be an analytic function in D and D
0
is a proper
subdomain, i.e.
D
0
D, with rectiﬁable boundary Γ. Then
Z
Γ
f(z)dz = 0.
31
In particular, if z
0
D
0
then
f(z
0
) =
1
2πi
Z
Γ
f(z)dz
z z
0
.
The Uniqueness Theorem. If f(z) and g(z) are analytic in D C and f(z
k
) = g(z
k
) for
some sequence {z
k
}
k=1
D which has an accumulation point in D, then f(z) g(z) in D .
The Maximum Principle. If f (z) is analytic in D C and continuous in the closure
D,
then for any subdomain U D one h olds
max
zU
|f(z)| = max
zU
|f(z)|,
where U denotes the boundary of U.
A holomorphic function f(z) is said to be entire (or integer ) if it is analytic in the whole
complex plane D = C. In other words, the entire functions are the largest class of functions
holomorphic in the ﬁnite plane which are the limit functions of convergent sequences of
polynomials, the convergence being uniform on every compact set.
Cauchy-Liouville Theorem. Let f(z) be a bounded entire function. Then f (z) = const.
2.1.2. Singular points. Let D be an open set. A point z
0
D of ﬁnite complex plane
is said to be an isolated singular point of an analytic function f(z) if f(z) is analytic in a
small punctured disk {z : 0 < |z z
0
| < ε} D and is unbounded there (otherwise, the
point z
0
is called regular and in that case f(z) can be continued up to an analytic function
at z
0
).
Let z
0
be an isolated singular point. Then f(z) can be represented by the Laurent series:
f(z) =
X
n=−∞
a
n
(z z
0
)
n
, (2.1)
where
a
n
=
1
2πi
Z
C
ρ
f(z)dz
(z z
0
)
n+1
, (2.2)
and C
ρ
is the circle centered at z
0
The following alternative is possible:
(i) if lim
zz
0
f(z) = then z
0
is called a pole of f(z); in this case a
n
= 0 for suﬃciently
large negative n < 0. In other words, there is a positive integer N N such that (zz
0
)
N
f(z)
is analytic in D. The smallest such an N is called the order of the pole z
0
. An other equivalent
deﬁnition is that z
0
is zero of 1/f(z) of order N.
(ii) if the limit lim
zz
0
f(z) does not exist then z
0
is called an essential singular point of
f(z).
A function f(z) is said to be meromorphic if it is holomorphic save for poles in the
ﬁnite complex plane. Typical examples of meromorphic functions is rational functions or
f(z) = tan z.
32
One of the main characteristics of the holomorphic function at an isolated singularity a
0
is the constant a
1
in the Laurent series (2.1). This coeﬃcient is called the residue of f(z)
0
. The residue of a function f around a point z
0
is also deﬁned by
res
z=z
0
f(z) =
1
2πi
Z
γ
f(z)dz,
where γ is counterclockwise simple closed contour, small enough to avoid any other poles of
f. Cauchy integral theorem implies that unless z
0
is a pole of f, its residue is zero.
Residue Theorem. If the contour γ encloses multiple poles a in domain D, then
Z
γ
f(z)dz = 2πi
X
res
z=a
f(z).
All the functions considered above are assumed to be single-valued, but sometimes we
will consider the inverse functions, which are, normally, inﬁnitely many valued . This leads
us to new types of singularities, such as algebraic and logarithmic branch points. We refer
an interested r eader to monograph .
2.2. Periods of analytic functions
2.2.1. Basic properties. In all of what follows, unless otherwise stated, we will assume
a function to be single-valued analytic function whose singularities do not have limit points
at the ﬁnite complex plane. If f (z) is such a function and if at each regular point z
f(z + Ω) = f(z),
where is a constant, then the number is called a period of f. Zero is a trivial period. A
function f(z) having nontrivial periods is said to be periodic. We denote by T (f ) the set of
all periods of f.
Proposition 2.1. If
1
, . . . ,
n
are peri ods of a function f, then for any integers
m
1
, . . . , m
n
the number
m
1
1
+ . . . + m
n
n
is also a period of f. In other words, T (f) is a module over Z.
The proof is an easy corollary of the deﬁnition.
Proposition 2.2. Let f(z) and g(z) have a p