COMPLEX ANALYSIS ASSIGNMENT 13
SIDDHARTH KOTHARI
Problem 1
Expanding the left side, we get
a b
e f
a b eτ + f
τ =
c d
g h
c d gτ + h
=
eτ +f
a gτ
+h + b
eτ +f
c gτ
+h + d
a(eτ + f ) + b(gτ + h)
c(eτ + f ) + d(gτ + h)
τ (ae + bg) + (af + bh)
=
τ (ce + dg) + (cf + dh)
ae + bg af + bh
=
τ
ce + dg cf + dh
a b
e f
=
τ,
c d
g h
=
as desired.
Problem 2
By the definition of the group action of SL2 (Z) on H we have
γτ :=
aτ + b
aτ + b cτ + d
aτ + b cτ + d
ac|τ |2 + adτ + bcτ + bd
=
=
=
,
cτ + d
cτ + d cτ + d
cτ + d cτ + d
|cτ + d|2
since the complex conjugate of an integer is itself. Therefore,
I(γτ ) =
1
I(adτ + bcτ ),
|cτ + d|2
and so I(γτ ) > 0 ⇐⇒ I(adτ + bcτ ) > 0. Let τ = α + βi with β > 0. Then
I(adτ + bcτ ) = I(ad(α + βi) + bc(α − βi)) = I(α(ad + bc) + βi(ad − bc)) = β(ad − bc) = β > 0,
which completes the proof.
Problem 3
For convenience, define
Rn := {z ∈ H : n ≤ R(z) < n + 1} .
Observe that
H=
[
Rn .
n∈Z
Now, let z ∈ H. By the above logic, we have that z ∈ Rn for a unique n. Note
n
1 1
z ∈ R0 ,
0 1
1
2
SIDDHARTH KOTHARI
n
1 1
as αn =
∈ SL2 (Z) sends τ ∈ H to τ + n. If αn z ∈ Ω, then we are done. Otherwise, we
0 1
have αn z ∈ R0 \ Ω, meaning that
1
1
z1 = α n z ∈ z ∈ H : z −
<
.
2
2
We will show that
1
1
1 1
0 −1
+m <
z2 =
z1 ̸∈ z ∈ H : z −
0 1
1 0
2
2
for all integers m. That is, we must show that
1
1
z2 −
+m ≥
2
2
for all integers m. Note that z2 = 1 − z11 and so
|z2 − (1/2 + m)| = (1/2)
z1 (1 + 2m) − 2
.
z1
Therefore, it suffices to show that |z1 (1 + 2m) − 2|2 ≥ |z1 |2 . Let z1 = a + bi where 0 < a < 1 and
b > 0. Thus, we must show
((1 + 2m)a − 2)2 + (1 − 2m)2 b2 ≥ a2 + b2 ,
and since (1 + 2m)2 b2 ≥ b2 , it suffices to show that ((1 + 2m)a − 2)2 ≥ a2 ⇐⇒ |(1 + 2m)a − 2| ≥ a
if and only if ((1 − 2m)a − 2) ≥ a ⇐⇒ m ≥ 1/a or ((1 − 2m)a − 2) ≤ −a ⇐⇒ m ≤ 1/a − 1
Problem 5
Let f : H → D = B1 (0) be a bijective holomorphic function, that is, f is a conformal mapping from
H to D. Since f is bijective (in particular, surjective), there exists a z0 ∈ H such that f (z0 ) = 0.
Define h : H → D by h(z) = e−iϕ f (z), where ϕ = arg(f ′ (z0 )).
Note that h is a conformal mapping from Ω = H to D such that h(z0 ) = 0 and h′ (z0 ) > 0 for
some z0 ∈ H. Thus, by the Riemann mapping theorem (as applied to H), such an h is unique.
for some β ∈ H. Let β = c + di with
Consider the function B : H → C defined by B(z) = i z−β
z−β
d > 0. Thus, for real b > 0, we have |b−d| < |b+d| =⇒ (b−d)2 < (b+d)2 =⇒ (a−c)2 +(b−d)2 <
(a − c)2 + (b + d)2 for all a, c ∈ R. Thus, |(a + bi) − (c + di)|2 < |(a + bi) − (c − di)|2 =⇒ |z − β|2 <
< 1, where z = a + bi is an arbitrary element in H. Therefore, the image of B
|z + β|2 =⇒ i z−β
z+β
is contained in D.
Next, note that
(z − β) − (z − β)
β−β
B ′ (z) = i
=i
,
2
(z − β)
(z − β)2
and so B is holomorphic on its domain (as β ̸∈ H), with a non-zero derivative. Furthermore,
i
i
i
observe that B ′ (β) = β−β
= (c+di)−(c−di)
= 2di
= 1/2d > 0, which is a positive real number, and
B(β) = 0.
Next, define g : D → C by g(z) := zβ−β
z−1 . Firstly, note that I(g(z)) > 0 for all z ∈ D, so the
image of g is contained in H. Secondly, g ′ (z) = β(z−1)−(βz−β)
, g is holomorphic on its domain (1
(z−1)2
is not in the unit disk). Most importantly, g ◦ B : H → H is the identity as is B ◦ g : D → D.
Therefore, B is bijective.
z−β
From our earlier analysis, all conformal maps from H to H are thus of the form α z−β
where α is
a unit complex number and β ∈ H, which is what we wanted to show.