SOLUTIONS TO HW6
Stein and Shakarchi, Chapter 5
Exercise 1. Let f (z) be a holomorphic function on the open unit disc D, and let {z1 , . . . , zN }
be the zeros of f (z) inside D (counted with multiplicity). For α ∈ D, we let ψα (z) denote the
function
α−z
ψα (z) =
.
1 − αz
This is similar to the function T (z) from Chapter 3, Problem 2.
Consider the function
f (z)
g(z) =
.
ψz1 (z) · ψz2 (z) · · · ψzN (z)
By construction, g(z) is nonvanishing on D. We proceed exactly as in Step 3 of the proof of
Jensen’s formula to conclude that
Z 2π
1
log(|g(eiθ )|) dθ.
log(|g(0)|) =
2π 0
We are left to consider the Blaschke factors. Since the factors in Jensen’s formula are
additive, we need only prove the formula for ψα (z) with α ∈ D. Since the only zero of ψα (z)
occurs at z = α, this boils down to showing
Z 2π
Z 2π
α − eiθ 1
1
iθ
dθ.
0=
log(|ψα (e )|) dθ =
log 2π 0
2π 0
1 − αeiθ Indeed, we have
1
2π
2π
Z
0
Z 2π
iθ
α − eiθ 1
1
|α
−
e
|
dθ =
log log
dθ
1 − αeiθ 2π 0
|eiθ | |e−iθ − α|
!
Z 2π
1
|α − eiθ |
=
log
dθ
2π 0
|α − eiθ |
Z 2π
1
=
log (1) dθ
2π 0
= 0.
Exercise 3. Let τ be a fixed complex number with =(τ ) > 0, and set
Θ(z|τ ) =
∞
X
πin2 τ 2πinz
e
e
n=−∞
=
∞
X
2 τ +2nz)
eπi(n
.
n=−∞
Write τ = σ + it, with t > 0. We then have
2 τ +2nz)
|eπi(n
| = e<(πin
2 σ−πn2 t+2πinx−2πny)
= eπ(−n
1
2 t−2ny)
2 t+2n|z|)
≤ eπ(−n
2 t/2
≤ e−πn
2
SOLUTIONS TO HW6
for n ≥ 4|z|/t. Note that this type of inequality also holds if n ≤ −4|z|/t. Therefore,
X
X
πi(n2 τ +2nz)
πi(n2 τ +2nz) |Θ(z|τ )| = e
+
e
|n|≤4|z|/t
|n|≥4|z|/t
X πi(n2 τ +2nz) X πi(n2 τ +2nz) +
e
e
≤ |n|≥4|z|/t
|n|≤4|z|/t
X
X
2
2
≤
eπ(−n t−2ny) +
e−πn t/2
|n|≤4|z|/t
≤ 2
|n|≥4|z|/t
−πn2 t 2πn|z|
X
e
e
+
≤ 2
e2πn|z| +
Z
?
≤
??
≤
2 t/2
e−πx
dx + 1
r
2
+1
t
0≤n≤4|z|/t
r
8|z|
2
2
+ 2 e8π|z| /t +
+1
t
t
!
r
8
2
2
2
|z|e8π|z| /t +
+ 3 e8π|z| /t
t
t
!
r
2
8 8π|z|2 /t+|z|
2
e
+
+ 3 e8π|z| /t+|z|
t
t
!
r
8
2
2
+
+ 3 e1 e(8π/t+1)|z| ,
t
t
≤ 2
≤
∞
−∞
0≤n≤4|z|/t
≤
2 t/2
e−πn
n=−∞
0≤n≤4|z|/t
X
∞
X
8π|z|2 /t
X
e
+
where in the inequality (?) we have used the fact that |z| ≤ e|z| , and for the inequality
2
(??) we have used that |z| ≤
p|z| + 1. Therefore, we see that Θ(z|τ ) is of order at most 2
(with constants A = (8/t + 2/t + 3)e and B = (8π/t + 1)). We should remark that these
computations can be done much less explicitly, but here we have instead opted for clarity.
To show that the order of Θ(z|τ ) is exactly 2, we exploit the symmetry of the definition of
Θ. Let α ∈ Z be an integer, and notice that, after reindexing the sum, we have
∞
X
2
Θ(ατ |τ ) =
eπiτ (n +2nα)
=
n=−∞
∞
X
eπiτ ((n+α)
2 −α2 )
n=−∞
= e
−πiτ α2
= e
−πiτ α2
∞
X
0 2
eπiτ (n )
n0 =−∞
Θ(0|τ ).
SOLUTIONS TO HW6
3
Now assume that Θ(z|τ ) was of order ρ < 2. We would then have (for some positive constants
A0 , B 0 )
0
ρ
A0 eB |z| ≥ |Θ(z|τ )|,
and in particular
0
A0 eB |τ |
ρ |α|ρ
2
2
≥ |Θ(ατ |τ )| = |e−πiτ α Θ(0|τ )| = |Θ(0|τ )|eπtα .
Taking the integer α sufficiently large gives us a contradiction.
Exercise 9. Let |z| < 1, and set
n
∞
Y
Y
k
2k
fn (z) =
(1 + z ),
f (z) =
(1 + z 2 ) = (1 + z)(1 + z 2 )(1 + z 4 )(1 + z 8 ) · · · .
k=0
k=0
P
2k
We first note that the infinite product for f (z) converges, since the sum ∞
converges.
k=0 |z|
Moreover, the sequence fn (z) converges to f (z) uniformly, by Proposition 3.2. We have
n
fn (z) = (1 + z)(1 + z 2 ) · · · (1 + z 2 )
n+1 −1
= 1 + a1 z + a2 z 2 + . . . + a2n+1 −1 z 2
,
and by computing the expansion of the product, we see that aj is the number of ways to write
the integer j as a sum
j = α0 + α1 21 + α2 22 + . . . + αn 2n ,
with αk ∈ {0, 1}; that is, aj is the number of binary expansions of j. Since binary expansions
are unique, we see that aj = 1 for every j. Hence
n+1
2
fn (z) = 1 + z + z + . . . + z
2n+1 −1
1 − z2
.
=
1−z
Note that we may also prove this by induction; evidently f0 (z) = 1 + z =
assume fn−1 (z) =
n
1−z 2
1−z
= 1 + z + ... + z
2n −1
1−z 2
,
1−z
and if we
, we have
n
n+1
1 − z2
1 − z2
n
fn (z) = fn−1 (z)(1 + z ) =
(1 + z 2 ) =
.
1−z
1−z
Using these results, we get
1
f (z) = lim fn (z) =
.
n→∞
1−z
2n
Exercise 10. (a) Let f (z) = ez − 1. One can easily check that the order of f (z) is 1.
The zeros of f (z) are exactly at the complex numbers of the form zk = 2πik, k ∈ Z. Since
f 0 (2πik) = e2πik = 1, we see that all of these zeros are simple. Therefore, Hadamard’s theorem
tells us that f (z) must be of the form
∞
∞ Y
Y
z z
z2
z
αz+β
αz+β
e 2πik = e
z
e − 1 = f (z) = e
z
1−
1+ 2 2
2πik
4π k
k=−∞,k6=0
k=1
for some α, β ∈ C. It remains to determine α and β. We have
∞ Y
ez − 1
z2
αz+β
=e
1+ 2 2 ,
z
4π k
k=1
4
SOLUTIONS TO HW6
which implies
∞ Y
z2
ez − 1
αz+β
1 + 2 2 = eβ .
1 = lim
= lim e
z→0
z→0
z
4π k
k=1
Therefore, we may take β = 0 (any other choice of β differs by a multiple of 2πi, but this
won’t affect the product). Expanding the first few terms of the product shows that we must
also have α = 21 and hence
∞ Y
z2
z
z/2
e −1=e z
1+ 2 2 .
4π k
k=1
(b) Now let f (z) = cos(πz). As in Example 1 in Section 3, f (z) has order equal to 1.
2k+1
The zeros of f (z) are
exactly
at complex numbers of the form zk = 2 , with k ∈ Z. Since
f 0 2k+1
= −π sin (2k+1)π
= ±π, all of the zeros are simple. Hadamard’s theorem again
2
2
gives
∞ ∞ 2
Y
Y
2z
2z
4z
αz+β
αz+β
cos(πz) = f (z) = e
e 2k+1 = e
1−
1−
2k
+
1
(2k + 1)2
k=−∞
k=0
for some α, β ∈ C. We again compute α and β:
∞ Y
αz+β
1 = lim cos(πz) = lim e
1−
z→0
z→0
k=0
4z 2
(2k + 1)2
= eβ
which implies β = 0. To find α, we simply note that cos(πz) is even; therefore
∞ ∞ Y
Y
4z 2
4z 2
αz
−αz
e
= cos(πz) = cos(−πz) = e
,
1−
1−
(2k + 1)2
(2k + 1)2
k=0
k=0
which implies α = 0. Therefore
∞ Y
cos(πz) =
1−
k=0
4z 2
(2k + 1)2
.
Exercise 13. Let f (z) = ez − z. One can easily check that f (z) has order 1. Assume that
ez − z = 0 has finitely many solutions, meaning f (z) has finitely many roots a1 , . . . aN . Each
of these roots is simple, and none are are equal to 0. Hadamard’s theorem gives us
N Y
z
z
z
z
γz+β
z
αz+β
ak
e =e
1−
··· 1 −
e − z = f (z) = e
1−
ak
a1
aN
k=1
P
−1
where α, β ∈ C, and where we denote γ = α + N
for brevity. This implies that
k=1 ak
(1−γ)z
−γz
th
e
− ze
is a polynomial, so that if ` > N , its ` coefficient is 0. If γ = 0, we
immediately obtain a contradiction, so we may assume that γ 6= 0. This gives
`
(1 − γ)`
(−γ)`
1
`
`
−
= 0 ⇐⇒ (γ − 1) − `γ = 0 ⇐⇒ 1 −
=`
`!
(` − 1)!
γ
for all ` > N . Since exponential functions grow (or decay) much faster than polynomials, we
obtain a contradiction.
SOLUTIONS TO HW6
5
Exercise 15. Let f (z) be a nonconstant meromorphic function on C, and let a1 , . . . , aN , . . . be
the (possibly infinite) set of poles of f (z), counted with multiplicity. We note that these poles
must necessarily be isolated by definition of meromorphicity, and we have limn→∞ an = ∞
(after potentially renumbering the an ), for otherwise our function f (z) would be constant. By
Theorem 4.1, there exists an entire function g(z) that vanishes exactly at the points {an }n≥1
and nowhere else. Therefore, the function
h(z) = f (z)g(z)
has removable singularities at all the points an , and therefore is entire. Hence
f (z) =
h(z)
g(z)
is a quotient of two entire functions.
Now assume a = {an }n≥1 and b = {bn }n≥1 be two disjoint sequences having no finite limit
points (this implies in particular that we may rearrange the sequences {an }n≥1 and {bn }n≥1
so that limn→∞ an = limn→∞ bn = ∞). Then, by Theorem 4.1, there exist entire functions
fa (z) and fb (z) which vanish exactly at {an }n≥1 and {bn }n≥1 , respectively. Therefore, the
function
fa (z)
fb (z)
has zeros precisely at {an }n≥1 and poles precisely at {bn }n≥1 .
Problem 1. Let f (z) be holomorphic on D, bounded, and not identically 0. Let z1 , z2 , . . .
be its zeros. We may assume that f (0) 6= 0. Fix R < 1 such that R 6= |zn | for every n.
Then f (z) must have finitely many zeros on DR (0) (else f (z) would be identically 0). After
renumbering the zeros so that |z1 | ≤ |z2 | ≤ . . ., we may assume these zeros contained in
DR (0) are z1 , z2 , . . . , zNR . Note that NR approaches ∞ as R approaches 1 from the left.
We’ll give two proofs of the statement using different techniques (but of the same flavor).
First solution. Let |f (z)| ≤ M on D, and let 0 < r ≤ R < 1. By Jensen’s fomula, we have
NR
Nr
X
X
R
R
log
log
≤
|z
|zk |
k|
k=1
k=1
Z 2π
1
=
log(|f (Reiθ )|) dθ − log(|f (0)|)
2π 0
≤ log(M ) − log(|f (0)|)
M
= log
.
|f (0)|
Letting R approach 1 from the left, and using the inequality 1 − x ≤ − log(x) for 0 < x < 1,
we obtain
Nr
Nr
X
X
M
1
1 − |zk | ≤
log
≤ log
.
|z
|f
(0)|
k|
k=1
k=1
Finally, letting r approach 1 from the left gives
∞
X
M
1 − |zk | ≤ log
< ∞.
|f
(0)|
k=1
6
SOLUTIONS TO HW6
Second solution. Assume the statement fails; that is, assume
n
X
lim
1 − |zk | = ∞.
n→∞
k=1
Since 1 − x ≤ − log(x) for 0 < x < 1, we obtain
∞ = lim
n→∞
n
X
k=1
1 − |zk | ≤ lim
n→∞
n
X
− log(|zk |) = lim − log
n→∞
k=1
n
Y
!
|zk | ,
k=1
which implies
lim
n→∞
n
Y
|zk | = 0.
k=1
Now let |f (z)| ≤ M on D, let 0 < r ≤ R < 1, and let k = 1, 2, . . . , Nr . We let z be of
modulus R, and write z = Reiθ . We then have
|R(zk − z)| = |Rzk − R2 eiθ | = |eiθ ||Re−iθ zk − R2 | = |R2 − zk z| = |R2 − zk z|.
Now set
Nr
Y
R2 − zk z
R(zk − z)
k=1
fR,r (z) = f (z) ·
!
;
the computation above implies |fR,r (z)| = |f (z)| for |z| = R. By the Maximum modulus
principle, we have
!
Nr
Y
RNr |f (0)| =
|zk | |fR,r (0)|
k=1
≤
Nr
Y
!
|zk | max(|fR,r (z)|)
|z|=R
k=1
=
Nr
Y
!
|zk | max(|f (z)|)
|z|=R
k=1
≤
Nr
Y
!
|zk | M.
k=1
Letting R approach 1 from the left gives
|f (0)| ≤
Nr
Y
!
|zk | M.
k=1
Letting r approach 1 from the left shows that
|f (0)| ≤ lim−
r→1
which contradicts the assumption f (0) 6= 0.
Nr
Y
k=1
!
|zk | M = 0,