SOLUTIONS OF HW9
MINGFENG ZHAO
April 23, 2011
1. [Problem 4, in Page 109] Let f (z) be a holomorphic function in the whole plane, and suppose that
there is an integer n and two positive real numbers R and M such that for all |z| ≥ R, we have
|f (z)| ≤ M |z|n .
Show then that f (z) is a polynomial of degree at most n.
Proof. For any z ∈ C and fix z. Take any r > R, then we know |z| < |z| + r.
Figure 1. |z| < |z| + r
Since f (z) is entire function, by Cauchy’s integral formula, we have
f
(n+1)
(n + 1)!
(z) =
2πi
Z
|ξ|=|z|+r
1
f (ξ)
dξ
(ξ − z)n+2
2
MINGFENG ZHAO
So we get
(n+1) (z) =
f
≤
≤
=
≤
≤
=
=
=
(n + 1)! Z
f (ξ)
dξ
n+2
2πi
(ξ
−
z)
|ξ|=|z|+r
Z
(n + 1)!
|f (ξ)|
|dξ|
2π
|ξ
− z|n+2
|ξ|=|z|+r
Z
(n + 1)!
|M |ξ|n |
|dξ| Since |ξ| = |z| + r > R
n+2
2π
|ξ|=|z|+r |ξ − z|
Z
M (n + 1)!
(|z| + r)n
|dξ|
n+2
2π
|ξ|=|z|+r |ξ − z|
Z
M (n + 1)!
(|z| + r)n
|dξ|
n+2
2π
|ξ|=|z|+r |ξ − z|
Z
(|z| + r)n
M (n + 1)!
|dξ| Since |ξ − z| ≥ |ξ| − |z|
n+2
2π
|ξ|=|z|+r (|ξ| − |z|)
Z
M (n + 1)!
(|z| + r)n
|dξ|
n+2
2π
|ξ|=|z|+r (|z| + r − |z|)
Z
M (n + 1)!
(|z| + r)n
|dξ|
2π
rn+2
|ξ|=|z|+r
Z
M (n + 1)! (|z| + r)n
·
|dξ|
2π
rn+2
|ξ|=|z|+r
=
M (n + 1)! (|z| + r)n
·
· 2π(|z| + r)
2π
rn+2
=
M (n + 1)!
(|z| + r)n+1
rn+2
Taking r → ∞, we get
(n+1) (z) = 0
f
By the arbitrary of z ∈ C, we know f (n+1) (z) ≡ 0 in C, which implies that f (z) is a polynomial of
degree at most n.
2. [Problem 5, in Page 109] Let f be a non-constant, holomorphic function in a connected open set
D, and let D0 be a connected open set whose closure D0 is a compact subset of D. Show that, if |f (z)|
is a constant function on the boundary of D0 , then there is at least one zero of f (z) in D0 .
SOLUTIONS OF HW9
3
Proof. We assume |f (z)| = c ≥ 0 for all z ∈ ∂D0 . If c = 0, since D0 is a bounded domain, then by the
maximum module principle, we know for all z ∈ D0 , we have |f (z)| ≤ 0, which implies that f (z) = 0.
By the arbitrary of z ∈ D0 , so f (z) ≡ 0 in D0 . Since D is connected, and D0 ⊂ D, and f (z) ≡ 0 in
D0 , then by the identity theorem, we now f (z) ≡ 0 in D, contradiction.
Figure 2. D0 ⊂ D
So now assume c > 0. If there is no zero point of f (z) in D0 , which implies
1 D0 . Also for z ∈ ∂D0 , we know f (z)
=
1
|f (z)|
g(z) =
=
1
c
1
f (z)
is well define in
> 0, so we can define
1
,
f (z)
for all z ∈ D0
So g(z) is analytic in a neighborhood of D0 in D. Since D0 is compact, then there exists some
z0 ∈ D0 such that
0 < |g(z0 )| =
max |g(z)|
z∈D 0
1 = max z∈D 0 f (z)
=
max
z∈D 0
1
|f (z)|
4
MINGFENG ZHAO
If z0 ∈ D0 , since D0 is a bounded domain, then by the maximum module principle, we know that
g(z) is a constant function on D0 , so f (z) is a constant function on D0 . Since D is connected, and
D0 ⊂ D, and f (z) ≡ 0 in D0 , then by the identity theorem, we now f (z) ≡ 0 in D, contradiction.
It z0 ∈ ∂D0 , by the argument before, we know for any z ∈ D0 , we have |g(z)| < |g(z0 )|, that is
1
1
<
|f (z)|
|f (z0 )|
So we get for all z ∈ D0 , we have c = |f (z0 )| < |f (z)|, which contradicts with the maximum module
principle.
Therefore, we know that there is at least one zero of f (z) in D0 .
3.
[Problem 6, in Page 110] Let D be a bounded, connected, open set and consider n points
P1 , P2 , · · · , Pn in the plane R2 . Show that the product P P1 · P P2 · · · P Pn of the distances from
a point P ∈ R2 , which varies in the closure D, to the points P1 , P2 , · · · , Pn attains its maximum at a
boundary point of D.
Proof. Let P1 = z1 , P2 = z2 , · · · , Pn = zn , P = z ∈ C, consider
f (z) = (z − z1 )(z − z2 ) · · · (z − zn ),
z ∈ D.
Notice
P P1 · P P2 · · · P Pn = |(z − z1 )(z − z2 ) · · · (z − zn )| = |f (z)|
On the other hand, since f (z) is analytic in D, and f (z) is not constant function, and also D is
connected, then by the Maximum module principle, we know there exists some z0 ∈ ∂D such that
|f (z0 )| = max |f (z)|
z∈D
Which implies that the product P P1 · P P2 · · · P Pn of the distances from a point P ∈ R2 , which
varies in the closure D, to the points P1 , P2 , · · · , Pn attains its maximum at a boundary point of D.
SOLUTIONS OF HW9
5
4. [Problem 7, in Page 110] Let f (z) be a holomorphic function in the disc |z| < R, and put
M (r) = sup|z|=r |f (z)| for 0 ≤ r < R. Show that
a. M (r) is a continuous, monotonic increasing function of r in 0 ≤ r < R;
b. If f (z) is not constant, then M (r) is strictly increasing.
Proof. Case I: If f (z) ≡ a for all |z| < R for some a ∈ C, then we have
M (r) = sup |f (z)| = sup |a| = |a| for all 0 ≤ r < R
|z|=r
|z|=r
Hence M (r) is a continuous, monotonic increasing function of r in 0 ≤ r < R.
Case II: f (z) is not a constant function in |z| < R.
Since f (z) is holomorphic function in |z| < R, in particular f (z) is continuous in |z| < R. Then for
any 0 ≤ r < R, we know |f (z)| is continuous on |z| ≤ r, so the compactness of |z| ≤ r implies that
there exists some zr in |z| ≤ r such that
M (r) = |f (zr )|.
Also by the maximum module principle, and f (z) is not constant in |z| ≤ r, we know that |zr | = r.
So for any 0 ≤ r1 < r2 < R, then there exist some zr1 such that |zr1 | = r1 < r2 such that
M (r1 ) = |f (zr1 )|.
For f (z) in |z| < r2 , by the strong maximum principle, and f (z) is not constant in |z| < R, we
know that
M (r2 ) = sup |f (z)| > |f (zr1 || = M (r1 ).
|z|=r2
Therefore, M (r) is strictly increasing.
Claim: M (r) is continuous in 0 ≤ r < R.
6
MINGFENG ZHAO
Suppose not, that is, there exists some 0 ≤ r0 < R such that M (r) is not continuous at r0 . So
there exists some 0 > 0 such that for any n ≥ 1, there exists 0 ≤ rn < R such that |rn − r0 | <
R−r0
2n ,
but
|M (rn ) − M (r0 )| ≥ 0 .
By the argument before, we know for any rn , there exists zn such that |zn | = rn , and
|f (zn )| = M (rn ).
We know |zn | ≤ rn ≤ r0 +
of |z| ≤
R+r0
2 ,
R−r0
2n
≤ r0 +
R−r0
2
=
R+r0
2
< R for all n ≥ 1, then by the compactness
there exists a subsequence {znk }∞
k=1 , and z0 ∈ C such that
lim znk = z0 .
k→∞
This implies that |z0 | = limk→∞ |znk | = limk→∞ rnk = r0 . Since |f (z)| is continuous in |z| < R,
we know
lim |f (znk )| = lim M (rnk ) = |f (z0 )|.
k→∞
k→∞
So we get
||f (z0 )| − M (r0 )| ≥ 0 .
For M (r0 ) = sup|z|=r0 |f (z)|, by the continuity of |f (z)| on |z| = r, then there exists wk such that
|wk | < r0 , and limk→∞ |wk | = r0 , and
lim |f (wk )| = M (r0 ).
k→∞
Since limk→∞ |znk | = limk→∞ rnk = r0 , then for any wk , there exists K ≥ 1 such that whenever
l ≥ K, we have |znl | ≥ |wk |. By the increasing of M (r), we know
M (rnl ) ≥ |f (wk )|
SOLUTIONS OF HW9
7
Taking l → ∞, so
|f (z0 )| ≥ |f (wk )|
Taking k → ∞, we get
|f (z0 )| ≥ M (r0 ).
On the other hand, since |z0 | = r0 , so |f (z0 )| ≤ M (r0 ) = sup|z|=r0 |f (z)|. Therefore, we know
M (r0 ) = |f (z0 )|, which contract its with ||f (z0 )| − M (r0 )| ≥ 0 . Therefore, M (r) is continuous in
0 ≤ r < R.
5. [Problem 12, in Page 166] Prove the following generalization of Schwarz’s lemma: If w(z) is analytic
and |w(z)| ≤ 1 for all |z| < 1, then for all |z| < 1, we have
|w(0)| − |z|
|w(0)| + |z|
≤ |w(z)| ≤
1 − |w(0)||z|
1 + |w(0)||z|
Proof. Case I: |w(0)| = 1.
Since |w(z)| ≤ 1 for all |z| < 1, then by the maximum module principle, we know w(z) ≡ w(0) for all
|z| < 1. Of course, we have
1=
|w(0)| + |z|
|w(0)| − |z|
≤ |w(z)| ≤
= 1.
1 − |w(0)||z|
1 + |w(0)||z|
Case II: |w(0)| < 1
Since |w(z)| ≤ 1 for all |z| < 1, then |w(0)||w(z)| < 1. So we can define
f (z) =
w(z) − w(0)
1 − w(0)w(z)
,
|z| < 1.
So we get f is analytic in |z| < 1, and |f (z)| ≤ 1, and f (0) = 0. By the Schwartz’s lemma, we get
w(z) − w(0) |f (z)| = ≤ |z|,
1 − w(0)w(z) |z| < 1.
8
MINGFENG ZHAO
On the other hand, We can find that
|w(0)| − |z|
≤ |w(z)|
1 − |w(0)||z|
⇐⇒
|w(0)| − |z| ≤ |w(z)| − |w(0)||z|
⇐⇒
|w(0)| − |w(z)| ≤ |z| − |w(0)||w(z)||z|
⇐⇒
|w(0)| − |w(z)| ≤ (1 − |w(0)||w(z)|)|z|
⇐⇒
|w(0)| − |w(z)|
≤ |z|
1 − |w(0)||w(z)|
⇐⇒
|w(0)| + |z| ≥ |w(z)| + |w(0)||w(z)||z|
⇐⇒
|z| − |w(0)||w(z)||z| ≥ |w(z)| − |z|
⇐⇒
(1 − |w(0)||w(z)|)|z| ≥ |w(z)| − |z|
⇐⇒
|z| ≥
And also, we can find that
|w(0)| + |z|
≥ |w(z)|
1 + |w(0)||z|
|w(z)| − |w(0)|
1 − |w(0)||w(z)|
So we have
|w(0)| − |w(z)| |w(0)| − |z|
|w(0)| + |z|
≤ |z|
≤ |w(z)| ≤
⇐⇒ 1 − |w(0)||z|
1 + |w(0)||z|
1 − |w(0)||w(z)| |w(0)|−|w(z)| w(z)−w(0) So it suffices to verify that 1−|w(0)||w(z)|
≤ 1−w(0)w(z) for all |z| < 1.
6. [Problem 13, in Page 166] The function f (z) and g(z) have poles of orders m and n, respectively,
at the point z = a. What can one say about the function f (z) + g(z), f (z)g(z) and f (z)/g(z) at this
point?
Proof. Since f (z) and g(z) have poles of orders m and n, respectively, at the point z = a, so we can
assume in a neighborhood |z − a| < r of z = a, there exists analytic functions F, G on |z − a| < r such
SOLUTIONS OF HW9
9
that F (a) 6= 0, and G(a) 6= 0, and
f (z) =
F (z)
,
(z − a)m
and g(z) =
G(z)
(z − a)n
Case I: m > n.
Ifm > n, then m − n > 0. So we can get
f (z) + g(z)
f (z)g(z)
f (z)
g(z)
=
F (z)
G(z)
+
(z − a)m
(z − a)n
=
F (z) + (z − a)n−m G(z)
(z − a)m
=
G(z)
F (z)
m
(z − a) (z − a)n
=
F (z)G(z)
(z − a)m+n
=
F (z)
(z − a)n
·
m
(z − a)
G(z)
=
(z − a)m−n
F (z)
G(z)
F (z)
But we know F (z) + (z − a)n−m G(z), F (z)G(z) and (z − a)m−n G(z)
are analytic in |z − a| < r, and
also F (z) + (z − a)n−m G(z), F (z)G(z) are equal to 0 at z = a.
So z = a is a ple of order m of f (z) + g(z), z = a is a pole of order m + n of f (z)g(z), and z = a is
a removable pole of
f (z)
g(z) .
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu