Maximum Modulus Principle - Complex Analysis, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


Chapter 3: The maximum modulus principle
Theorem 3.1 (Identity theorem for analytic functions) Let G  C be open and connected
(and nonempty). Letf : G!C be analytic. Then the following are equivalent forf:
(i) f 0
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) = 08n
(iii) there is a pointa2G withf
(n)
(a) = 0 forn = 0; 1; 2;:::.
Proof. We show (i)) (ii)) (iii)) (i).
(i)) (ii): is really obvious. Iff 0, take anya2 G (here we needG6=;), choose > 0
withD(a;)G and putz
n
=a +=(n + 1).
(ii)) (iii): Assuming lim
n!1
z
n
=a2G,z
n
distinct andf(z
n
) = 08n, consider the power
series forf centered ata. That is
f(z) =
1
X
n=0
a
n
(za)
n
forjzaj<
(for some> 0 withD(a;)G). Herea
n
=f
(n)
(a)=n! and so our aim of showingf
(n)
(a) =
0 for alln 0 is equivalent to showinga
n
= 0 for alln. If that is not the case, there must be a
smallestm 0 witha
m
6= 0.
Now, forjzaj< we can write
f(z) =
1
X
n=m
a
n
(za)
n
= (za)
m
1
X
n=m
a
n
(za)
nm
= (za)
m
g(z)
1
Page 2


Chapter 3: The maximum modulus principle
Theorem 3.1 (Identity theorem for analytic functions) Let G  C be open and connected
(and nonempty). Letf : G!C be analytic. Then the following are equivalent forf:
(i) f 0
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) = 08n
(iii) there is a pointa2G withf
(n)
(a) = 0 forn = 0; 1; 2;:::.
Proof. We show (i)) (ii)) (iii)) (i).
(i)) (ii): is really obvious. Iff 0, take anya2 G (here we needG6=;), choose > 0
withD(a;)G and putz
n
=a +=(n + 1).
(ii)) (iii): Assuming lim
n!1
z
n
=a2G,z
n
distinct andf(z
n
) = 08n, consider the power
series forf centered ata. That is
f(z) =
1
X
n=0
a
n
(za)
n
forjzaj<
(for some> 0 withD(a;)G). Herea
n
=f
(n)
(a)=n! and so our aim of showingf
(n)
(a) =
0 for alln 0 is equivalent to showinga
n
= 0 for alln. If that is not the case, there must be a
smallestm 0 witha
m
6= 0.
Now, forjzaj< we can write
f(z) =
1
X
n=m
a
n
(za)
n
= (za)
m
1
X
n=m
a
n
(za)
nm
= (za)
m
g(z)
1
andg(z) =
P
1
n=m
a
n
(za)
nm
is analytic forjzaj<. Moreoverg(a) =a
m
6= 0,g(z) is
continuous atz =a and so we can ?nd
0
> 0,
0
, with
jg(z)g(a)j 
1
2
jg(a)j forjzaj<
0
)jg(z)j 
1
2
jg(a)j forjzaj<
0
)g(z) 6= 0 forjzaj<
0
:
But 0 = f(z
n
) and for n large enough (say n > N) we havejz
n
aj < 
0
so that f(z
n
) =
(z
n
a)
m
g(z
n
) = 0. Thus (forn>N)z
n
= 0 org(z
n
) = 0. However, we knowg(z
n
)6= 0 and
thez
n
are distinct so that at most onen can havez
n
=a. Hence we are faced with a contradiction.
The contradiction arose from assuming that there was anya
n
6= 0. We must therefore have
a
n
= 08n.
(iii)) (i): Assume now that there isa2G withf
(n)
(a) = 0 for alln 0. Then the power
series expansion forf abouta (which is valid in a discD(a;)G with> 0) is
f(z) =
1
X
n=0
f
(n)
(a)
n!
(za)
n
= 0 forjzaj<:
Thusf(z) 0 forjzaj< and differentiating we getf
(n)
(z) = 0 forn = 0; 1; 2;:::.
This shows thatU =fa2 G : f
(n)
(a) = 0 for alln = 0; 1; 2;:::g is open (and nonempty).
U is also closed relative toG. To see that takeb2GnU. Then there is somen withf
(n)
(b)6= 0.
Now that f
(n)
is continuous at b and so there is a  > 0 so that f
(n)
(z) 6= 0 for all z with
jzbj < . This means none of thesez can be inU or in other wordsD(b;) GnU. This
meansGnU is open.
AsG is connected,UG nonempty and both open and closed relative toG impliesU =G.
This meansf
(n)
(z) = 0 for alln 0 and allz2G. Speci?cally withn = 0 we havef 0.
Corollary 3.2 (version with two functions) LetGC be open and connected (and nonempty).
Letf;g : G!C be two analytic functions. Then the following are equivalent forf andg:
(i) fg
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) =g(z
n
)8n
(iii) there is a pointa2G withf
(n)
(a) =g
(n)
(a) forn = 0; 1; 2;:::.
Proof. apply the Identity Theorem 3.1 to the differencefg.
Remark 3.3 The signi?cance of the Identity Theorem is that an analytic function on a connected
openGC is determined on all ofG by its behaviour near a single point.
Thus if an analytic function is given on one part ofG by a formula likef(z) =
1
z1
and that
formula makes sense and gives an analytic function on a larger connected subset ofG then it has
to be thatf(z) =
1
z1
also holds in the larger set.
Page 3


Chapter 3: The maximum modulus principle
Theorem 3.1 (Identity theorem for analytic functions) Let G  C be open and connected
(and nonempty). Letf : G!C be analytic. Then the following are equivalent forf:
(i) f 0
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) = 08n
(iii) there is a pointa2G withf
(n)
(a) = 0 forn = 0; 1; 2;:::.
Proof. We show (i)) (ii)) (iii)) (i).
(i)) (ii): is really obvious. Iff 0, take anya2 G (here we needG6=;), choose > 0
withD(a;)G and putz
n
=a +=(n + 1).
(ii)) (iii): Assuming lim
n!1
z
n
=a2G,z
n
distinct andf(z
n
) = 08n, consider the power
series forf centered ata. That is
f(z) =
1
X
n=0
a
n
(za)
n
forjzaj<
(for some> 0 withD(a;)G). Herea
n
=f
(n)
(a)=n! and so our aim of showingf
(n)
(a) =
0 for alln 0 is equivalent to showinga
n
= 0 for alln. If that is not the case, there must be a
smallestm 0 witha
m
6= 0.
Now, forjzaj< we can write
f(z) =
1
X
n=m
a
n
(za)
n
= (za)
m
1
X
n=m
a
n
(za)
nm
= (za)
m
g(z)
1
andg(z) =
P
1
n=m
a
n
(za)
nm
is analytic forjzaj<. Moreoverg(a) =a
m
6= 0,g(z) is
continuous atz =a and so we can ?nd
0
> 0,
0
, with
jg(z)g(a)j 
1
2
jg(a)j forjzaj<
0
)jg(z)j 
1
2
jg(a)j forjzaj<
0
)g(z) 6= 0 forjzaj<
0
:
But 0 = f(z
n
) and for n large enough (say n > N) we havejz
n
aj < 
0
so that f(z
n
) =
(z
n
a)
m
g(z
n
) = 0. Thus (forn>N)z
n
= 0 org(z
n
) = 0. However, we knowg(z
n
)6= 0 and
thez
n
are distinct so that at most onen can havez
n
=a. Hence we are faced with a contradiction.
The contradiction arose from assuming that there was anya
n
6= 0. We must therefore have
a
n
= 08n.
(iii)) (i): Assume now that there isa2G withf
(n)
(a) = 0 for alln 0. Then the power
series expansion forf abouta (which is valid in a discD(a;)G with> 0) is
f(z) =
1
X
n=0
f
(n)
(a)
n!
(za)
n
= 0 forjzaj<:
Thusf(z) 0 forjzaj< and differentiating we getf
(n)
(z) = 0 forn = 0; 1; 2;:::.
This shows thatU =fa2 G : f
(n)
(a) = 0 for alln = 0; 1; 2;:::g is open (and nonempty).
U is also closed relative toG. To see that takeb2GnU. Then there is somen withf
(n)
(b)6= 0.
Now that f
(n)
is continuous at b and so there is a  > 0 so that f
(n)
(z) 6= 0 for all z with
jzbj < . This means none of thesez can be inU or in other wordsD(b;) GnU. This
meansGnU is open.
AsG is connected,UG nonempty and both open and closed relative toG impliesU =G.
This meansf
(n)
(z) = 0 for alln 0 and allz2G. Speci?cally withn = 0 we havef 0.
Corollary 3.2 (version with two functions) LetGC be open and connected (and nonempty).
Letf;g : G!C be two analytic functions. Then the following are equivalent forf andg:
(i) fg
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) =g(z
n
)8n
(iii) there is a pointa2G withf
(n)
(a) =g
(n)
(a) forn = 0; 1; 2;:::.
Proof. apply the Identity Theorem 3.1 to the differencefg.
Remark 3.3 The signi?cance of the Identity Theorem is that an analytic function on a connected
openGC is determined on all ofG by its behaviour near a single point.
Thus if an analytic function is given on one part ofG by a formula likef(z) =
1
z1
and that
formula makes sense and gives an analytic function on a larger connected subset ofG then it has
to be thatf(z) =
1
z1
also holds in the larger set.
This is quite different from what happens with continuous functions likef :C!C de?ned
by
f(z) =

z jzj< 1
z
jzj
jzj 1:
Even for C
1
functions we can have different formulae holding in different places. Consider
g :C!C where
f(z) =
8
<
:
0 jzj 1
exp



1
jzj1

2

jzj> 1:
The original meaning of the word ‘analytic’ related to this property of analytic functions (one
formula).
Corollary 3.4 IfGC is a connected open set andf : G!C is analytic and not identically
constant, then the zero set off
Z
f
=fz2G :f(z) = 0g
has no accumulation points inG.
Proof. First we should de?ne accumulation point in case you forget it. IfSC is any set and
a2C, thena is called an accumulation point ofS if for each> 0
(Snfag)\D(a;)6=;:
If wea is an accumulation point ofS we can choose
z
1
2 (Snfag)\D(a; 1)
z
2
2 (Snfag)\D

a; min

1
2
;jz
1
aj

z
3
2 (Snfag)\D

a; min

1
3
;jz
2
aj

and (inductively)z
n+1
2 (Snfag)\D

a; min

1
n
;jz
n
aj



. This produces a sequence (z
n
)
1
n=1
of distinct pointsz
n
2 S with lim
n!1
z
n
= a. (It is not hard to see that the existence of such a
sequence is equivalent toa being an accumulation point ofS.)
Applying this toS =Z
f
and using Theorem 3.1 we getf 0.
Corollary 3.5 LetGC be open and connected and letKG be compact. Letf;g : G!C
be analytic. If the equationf(z) =g(z) has in?nitely many solutionsz2K, thenfg.
Proof. Choose an in?nite sequence (z
n
)
1
n=1
of distinct points z
n
2 K where f(z
n
) = g(z
n
).
Since K is compact, the sequence has a convergent subsequence (z
n
j
)
1
j=1
with a limit a =
lim
j!1
z
n
j
2KG.
By Corollary 3.2,fg.
Page 4


Chapter 3: The maximum modulus principle
Theorem 3.1 (Identity theorem for analytic functions) Let G  C be open and connected
(and nonempty). Letf : G!C be analytic. Then the following are equivalent forf:
(i) f 0
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) = 08n
(iii) there is a pointa2G withf
(n)
(a) = 0 forn = 0; 1; 2;:::.
Proof. We show (i)) (ii)) (iii)) (i).
(i)) (ii): is really obvious. Iff 0, take anya2 G (here we needG6=;), choose > 0
withD(a;)G and putz
n
=a +=(n + 1).
(ii)) (iii): Assuming lim
n!1
z
n
=a2G,z
n
distinct andf(z
n
) = 08n, consider the power
series forf centered ata. That is
f(z) =
1
X
n=0
a
n
(za)
n
forjzaj<
(for some> 0 withD(a;)G). Herea
n
=f
(n)
(a)=n! and so our aim of showingf
(n)
(a) =
0 for alln 0 is equivalent to showinga
n
= 0 for alln. If that is not the case, there must be a
smallestm 0 witha
m
6= 0.
Now, forjzaj< we can write
f(z) =
1
X
n=m
a
n
(za)
n
= (za)
m
1
X
n=m
a
n
(za)
nm
= (za)
m
g(z)
1
andg(z) =
P
1
n=m
a
n
(za)
nm
is analytic forjzaj<. Moreoverg(a) =a
m
6= 0,g(z) is
continuous atz =a and so we can ?nd
0
> 0,
0
, with
jg(z)g(a)j 
1
2
jg(a)j forjzaj<
0
)jg(z)j 
1
2
jg(a)j forjzaj<
0
)g(z) 6= 0 forjzaj<
0
:
But 0 = f(z
n
) and for n large enough (say n > N) we havejz
n
aj < 
0
so that f(z
n
) =
(z
n
a)
m
g(z
n
) = 0. Thus (forn>N)z
n
= 0 org(z
n
) = 0. However, we knowg(z
n
)6= 0 and
thez
n
are distinct so that at most onen can havez
n
=a. Hence we are faced with a contradiction.
The contradiction arose from assuming that there was anya
n
6= 0. We must therefore have
a
n
= 08n.
(iii)) (i): Assume now that there isa2G withf
(n)
(a) = 0 for alln 0. Then the power
series expansion forf abouta (which is valid in a discD(a;)G with> 0) is
f(z) =
1
X
n=0
f
(n)
(a)
n!
(za)
n
= 0 forjzaj<:
Thusf(z) 0 forjzaj< and differentiating we getf
(n)
(z) = 0 forn = 0; 1; 2;:::.
This shows thatU =fa2 G : f
(n)
(a) = 0 for alln = 0; 1; 2;:::g is open (and nonempty).
U is also closed relative toG. To see that takeb2GnU. Then there is somen withf
(n)
(b)6= 0.
Now that f
(n)
is continuous at b and so there is a  > 0 so that f
(n)
(z) 6= 0 for all z with
jzbj < . This means none of thesez can be inU or in other wordsD(b;) GnU. This
meansGnU is open.
AsG is connected,UG nonempty and both open and closed relative toG impliesU =G.
This meansf
(n)
(z) = 0 for alln 0 and allz2G. Speci?cally withn = 0 we havef 0.
Corollary 3.2 (version with two functions) LetGC be open and connected (and nonempty).
Letf;g : G!C be two analytic functions. Then the following are equivalent forf andg:
(i) fg
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) =g(z
n
)8n
(iii) there is a pointa2G withf
(n)
(a) =g
(n)
(a) forn = 0; 1; 2;:::.
Proof. apply the Identity Theorem 3.1 to the differencefg.
Remark 3.3 The signi?cance of the Identity Theorem is that an analytic function on a connected
openGC is determined on all ofG by its behaviour near a single point.
Thus if an analytic function is given on one part ofG by a formula likef(z) =
1
z1
and that
formula makes sense and gives an analytic function on a larger connected subset ofG then it has
to be thatf(z) =
1
z1
also holds in the larger set.
This is quite different from what happens with continuous functions likef :C!C de?ned
by
f(z) =

z jzj< 1
z
jzj
jzj 1:
Even for C
1
functions we can have different formulae holding in different places. Consider
g :C!C where
f(z) =
8
<
:
0 jzj 1
exp



1
jzj1

2

jzj> 1:
The original meaning of the word ‘analytic’ related to this property of analytic functions (one
formula).
Corollary 3.4 IfGC is a connected open set andf : G!C is analytic and not identically
constant, then the zero set off
Z
f
=fz2G :f(z) = 0g
has no accumulation points inG.
Proof. First we should de?ne accumulation point in case you forget it. IfSC is any set and
a2C, thena is called an accumulation point ofS if for each> 0
(Snfag)\D(a;)6=;:
If wea is an accumulation point ofS we can choose
z
1
2 (Snfag)\D(a; 1)
z
2
2 (Snfag)\D

a; min

1
2
;jz
1
aj

z
3
2 (Snfag)\D

a; min

1
3
;jz
2
aj

and (inductively)z
n+1
2 (Snfag)\D

a; min

1
n
;jz
n
aj



. This produces a sequence (z
n
)
1
n=1
of distinct pointsz
n
2 S with lim
n!1
z
n
= a. (It is not hard to see that the existence of such a
sequence is equivalent toa being an accumulation point ofS.)
Applying this toS =Z
f
and using Theorem 3.1 we getf 0.
Corollary 3.5 LetGC be open and connected and letKG be compact. Letf;g : G!C
be analytic. If the equationf(z) =g(z) has in?nitely many solutionsz2K, thenfg.
Proof. Choose an in?nite sequence (z
n
)
1
n=1
of distinct points z
n
2 K where f(z
n
) = g(z
n
).
Since K is compact, the sequence has a convergent subsequence (z
n
j
)
1
j=1
with a limit a =
lim
j!1
z
n
j
2KG.
By Corollary 3.2,fg.
Theorem 3.6 (Maximum modulus theorem, basic version) Let G C be a connected open
set andf : G!C analytic. If there is anya2 G withjf(a)jjf(z)j for allz2 G, thenf is
constant.
Proof. (Another way to state this is thatjf(z)j cannot have a maximum inG, unlessf is con-
stant.)
Choose > 0 so thatD(a;)G. Fix 0<r< and then we have (by the Cauchy integral
formula)
f(a) =
1
2i
Z
jzaj=r
f(z)
za
dz:
Write this out in terms of a parametrisationz =a +re
i
with 0 2,dz =ire
i
d.
f(a) =
1
2i
Z
2
0
f(a +re
i
)
ire
i
d
1
2
Z
2
0
f(a +re
i
)d:
Hence
jf(a)j
1
2
Z
2
0
jf(a +re
i
)jd
1
2
Z
2
0
jf(a)jd =jf(a)j;
usingjf(a +re
i
)jjf(a)j8.
We must therefore have equality in the inequalities. Since the integrandjf(a +re
i
)j is a
continuous function of, this impliesjf(a +re
i
)j =jf(a)j for all.
Put = Arg(f(a)). Now
jf(a)j = e
i
f(a)
=
e
i
2
Z
2
0
f(a +re
i
)d
=
1
2
Z
2
0
e
i
f(a +re
i
)d
<jf(a)j =jf(a)j =
1
2
<
Z
2
0
e
i
f(a +re
i
)d
=
1
2
Z
2
0
<(e
i
f(a +re
i
))d

1
2
Z
2
0
je
i
f(a +re
i
)jd
using<wjwj forw2C

1
2
Z
2
0
jf(a +re
i
)jd

1
2
Z
2
0
jf(a)jd =jf(a)j
Thus we must again have equality in all the inequalities and so
<(e
i
f(a +re
i
)) =je
i
f(a +re
i
)j =jf(a)j
Page 5


Chapter 3: The maximum modulus principle
Theorem 3.1 (Identity theorem for analytic functions) Let G  C be open and connected
(and nonempty). Letf : G!C be analytic. Then the following are equivalent forf:
(i) f 0
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) = 08n
(iii) there is a pointa2G withf
(n)
(a) = 0 forn = 0; 1; 2;:::.
Proof. We show (i)) (ii)) (iii)) (i).
(i)) (ii): is really obvious. Iff 0, take anya2 G (here we needG6=;), choose > 0
withD(a;)G and putz
n
=a +=(n + 1).
(ii)) (iii): Assuming lim
n!1
z
n
=a2G,z
n
distinct andf(z
n
) = 08n, consider the power
series forf centered ata. That is
f(z) =
1
X
n=0
a
n
(za)
n
forjzaj<
(for some> 0 withD(a;)G). Herea
n
=f
(n)
(a)=n! and so our aim of showingf
(n)
(a) =
0 for alln 0 is equivalent to showinga
n
= 0 for alln. If that is not the case, there must be a
smallestm 0 witha
m
6= 0.
Now, forjzaj< we can write
f(z) =
1
X
n=m
a
n
(za)
n
= (za)
m
1
X
n=m
a
n
(za)
nm
= (za)
m
g(z)
1
andg(z) =
P
1
n=m
a
n
(za)
nm
is analytic forjzaj<. Moreoverg(a) =a
m
6= 0,g(z) is
continuous atz =a and so we can ?nd
0
> 0,
0
, with
jg(z)g(a)j 
1
2
jg(a)j forjzaj<
0
)jg(z)j 
1
2
jg(a)j forjzaj<
0
)g(z) 6= 0 forjzaj<
0
:
But 0 = f(z
n
) and for n large enough (say n > N) we havejz
n
aj < 
0
so that f(z
n
) =
(z
n
a)
m
g(z
n
) = 0. Thus (forn>N)z
n
= 0 org(z
n
) = 0. However, we knowg(z
n
)6= 0 and
thez
n
are distinct so that at most onen can havez
n
=a. Hence we are faced with a contradiction.
The contradiction arose from assuming that there was anya
n
6= 0. We must therefore have
a
n
= 08n.
(iii)) (i): Assume now that there isa2G withf
(n)
(a) = 0 for alln 0. Then the power
series expansion forf abouta (which is valid in a discD(a;)G with> 0) is
f(z) =
1
X
n=0
f
(n)
(a)
n!
(za)
n
= 0 forjzaj<:
Thusf(z) 0 forjzaj< and differentiating we getf
(n)
(z) = 0 forn = 0; 1; 2;:::.
This shows thatU =fa2 G : f
(n)
(a) = 0 for alln = 0; 1; 2;:::g is open (and nonempty).
U is also closed relative toG. To see that takeb2GnU. Then there is somen withf
(n)
(b)6= 0.
Now that f
(n)
is continuous at b and so there is a  > 0 so that f
(n)
(z) 6= 0 for all z with
jzbj < . This means none of thesez can be inU or in other wordsD(b;) GnU. This
meansGnU is open.
AsG is connected,UG nonempty and both open and closed relative toG impliesU =G.
This meansf
(n)
(z) = 0 for alln 0 and allz2G. Speci?cally withn = 0 we havef 0.
Corollary 3.2 (version with two functions) LetGC be open and connected (and nonempty).
Letf;g : G!C be two analytic functions. Then the following are equivalent forf andg:
(i) fg
(ii) there is an in?nite sequence (z
n
)
1
n=1
of distinct points ofG with lim
n!1
z
n
= a2 G and
f(z
n
) =g(z
n
)8n
(iii) there is a pointa2G withf
(n)
(a) =g
(n)
(a) forn = 0; 1; 2;:::.
Proof. apply the Identity Theorem 3.1 to the differencefg.
Remark 3.3 The signi?cance of the Identity Theorem is that an analytic function on a connected
openGC is determined on all ofG by its behaviour near a single point.
Thus if an analytic function is given on one part ofG by a formula likef(z) =
1
z1
and that
formula makes sense and gives an analytic function on a larger connected subset ofG then it has
to be thatf(z) =
1
z1
also holds in the larger set.
This is quite different from what happens with continuous functions likef :C!C de?ned
by
f(z) =

z jzj< 1
z
jzj
jzj 1:
Even for C
1
functions we can have different formulae holding in different places. Consider
g :C!C where
f(z) =
8
<
:
0 jzj 1
exp



1
jzj1

2

jzj> 1:
The original meaning of the word ‘analytic’ related to this property of analytic functions (one
formula).
Corollary 3.4 IfGC is a connected open set andf : G!C is analytic and not identically
constant, then the zero set off
Z
f
=fz2G :f(z) = 0g
has no accumulation points inG.
Proof. First we should de?ne accumulation point in case you forget it. IfSC is any set and
a2C, thena is called an accumulation point ofS if for each> 0
(Snfag)\D(a;)6=;:
If wea is an accumulation point ofS we can choose
z
1
2 (Snfag)\D(a; 1)
z
2
2 (Snfag)\D

a; min

1
2
;jz
1
aj

z
3
2 (Snfag)\D

a; min

1
3
;jz
2
aj

and (inductively)z
n+1
2 (Snfag)\D

a; min

1
n
;jz
n
aj



. This produces a sequence (z
n
)
1
n=1
of distinct pointsz
n
2 S with lim
n!1
z
n
= a. (It is not hard to see that the existence of such a
sequence is equivalent toa being an accumulation point ofS.)
Applying this toS =Z
f
and using Theorem 3.1 we getf 0.
Corollary 3.5 LetGC be open and connected and letKG be compact. Letf;g : G!C
be analytic. If the equationf(z) =g(z) has in?nitely many solutionsz2K, thenfg.
Proof. Choose an in?nite sequence (z
n
)
1
n=1
of distinct points z
n
2 K where f(z
n
) = g(z
n
).
Since K is compact, the sequence has a convergent subsequence (z
n
j
)
1
j=1
with a limit a =
lim
j!1
z
n
j
2KG.
By Corollary 3.2,fg.
Theorem 3.6 (Maximum modulus theorem, basic version) Let G C be a connected open
set andf : G!C analytic. If there is anya2 G withjf(a)jjf(z)j for allz2 G, thenf is
constant.
Proof. (Another way to state this is thatjf(z)j cannot have a maximum inG, unlessf is con-
stant.)
Choose > 0 so thatD(a;)G. Fix 0<r< and then we have (by the Cauchy integral
formula)
f(a) =
1
2i
Z
jzaj=r
f(z)
za
dz:
Write this out in terms of a parametrisationz =a +re
i
with 0 2,dz =ire
i
d.
f(a) =
1
2i
Z
2
0
f(a +re
i
)
ire
i
d
1
2
Z
2
0
f(a +re
i
)d:
Hence
jf(a)j
1
2
Z
2
0
jf(a +re
i
)jd
1
2
Z
2
0
jf(a)jd =jf(a)j;
usingjf(a +re
i
)jjf(a)j8.
We must therefore have equality in the inequalities. Since the integrandjf(a +re
i
)j is a
continuous function of, this impliesjf(a +re
i
)j =jf(a)j for all.
Put = Arg(f(a)). Now
jf(a)j = e
i
f(a)
=
e
i
2
Z
2
0
f(a +re
i
)d
=
1
2
Z
2
0
e
i
f(a +re
i
)d
<jf(a)j =jf(a)j =
1
2
<
Z
2
0
e
i
f(a +re
i
)d
=
1
2
Z
2
0
<(e
i
f(a +re
i
))d

1
2
Z
2
0
je
i
f(a +re
i
)jd
using<wjwj forw2C

1
2
Z
2
0
jf(a +re
i
)jd

1
2
Z
2
0
jf(a)jd =jf(a)j
Thus we must again have equality in all the inequalities and so
<(e
i
f(a +re
i
)) =je
i
f(a +re
i
)j =jf(a)j
for all. Thus=(e
i
f(a +re
i
)) = 0 ande
i
f(a +re
i
) =jf(a)j orf(a +re
i
) =e
i
jf(a)j.
Thus f(z) is constant for z in the in?nite compact subsetfz : jzaj = rg of G. By
Corollary 3.5, it follows thatf(z) is constant (onG).
Theorem 3.7 (Maximum modulus theorem, usual version) The absolute value of a noncon-
stant analytic function on a connected open setGC cannot have a local maximum point in
G.
Proof. Letf : G!C be analytic. By a local maximum point forjfj we mean a pointa2 G
wherejf(a)jjf(z)j holds for all z2 D(a;)\G, some > 0. AsG is open, by making
> 0 smaller if necessary we can assumeD(a;)G.
By Theorem 3.6,jf(a)jjf(z)j8z2 D(a;) impliesf constant onD(a;) (sincef must
be analytic on D(a;) G and D(a;) is connected open). Then, by the Identity Theorem
(Corollary 3.2),f must be constant.
Corollary 3.8 (Maximum modulus theorem, another usual version) LetGC be a bounded
and connected open set. Letf :

G!C be continuous on the closure

G ofG and analytic onG.
Then
sup
z2

G
jf(z)j = sup
z2@G
jf(z)j:
(That is the maximum modulus of the analytic functionf(z) is attained on the boundary@G.)
Proof. Since G is bounded, its closure

G is closed and bounded, hence compact. jf(z)j is a
continuous real-valued function on the compact set and so sup
z2

G
jf(z)j<1 and the supremum
is attained at some pointb2

G. That isjf(b)jjf(z)j8z2

G. Ifb is on the boundary@G then
we have
jf(b)j = sup
z2@G
jf(z)j = sup
z2

G
jf(z)j
but ifb2G, thenf must be constant by the Identity Theorem 3.1. So in that case sup
z2@G
jf(z)j =
sup
z2

G
jf(z)j is also true.
Theorem 3.9 (Fundamental theorem of algebra) Letp(z) be a nonconstant polynomial (p(z) =
P
n
k=0
a
k
z
k
witha
n
6= 0 andn 1). Then the equationp(z) = 0 has a solutionz2C.
Proof. Dividing the equation by the coef?cient of the highest power ofz with a nonzero coef?-
cient (a
n
in the notation above) we can assume without loss of generality that the polynomial is
monic (meaning that the coef?cient of the highest power ofz is 1) and that the polynomial is
p(z) =z
n
+
n1
X
k=0
a
k
z
k
(n 1):