Chapter 3: The maximum modulus principle
Course 414, 2003–04
December 3, 2003
Theorem 3.1 (Identity theorem for analytic functions) Let G ⊂ C be open and connected
(and nonempty). Let f : G → C be analytic. Then the following are equivalent for f :
(i) f ≡ 0
(ii) there is an infinite sequence (zn )∞
n=1 of distinct points of G with limn→∞ zn = a ∈ G and
f (zn ) = 0∀n
(iii) there is a point a ∈ G with f (n) (a) = 0 for n = 0, 1, 2, . . ..
Proof. We show (i) ⇒ (ii) ⇒ (iii) ⇒ (i).
(i) ⇒ (ii): is really obvious. If f ≡ 0, take any a ∈ G (here we need G 6= ∅), choose δ > 0
with D(a, δ) ⊂ G and put zn = a + δ/(n + 1).
(ii) ⇒ (iii): Assuming limn→∞ zn = a ∈ G, zn distinct and f (zn ) = 0∀n, consider the power
series for f centered at a. That is
f (z) =
∞
X
an (z − a)n for |z − a| < δ
n=0
(for some δ > 0 with D(a, δ) ⊂ G). Here an = f (n) (a)/n! and so our aim of showing f (n) (a) =
0 for all n ≥ 0 is equivalent to showing an = 0 for all n. If that is not the case, there must be a
smallest m ≥ 0 with am 6= 0.
Now, for |z − a| < δ we can write
f (z) =
∞
X
an (z − a)n
n=m
= (z − a)m
∞
X
n=m
= (z − a)m g(z)
1
an (z − a)n−m
2
414 2003–04 R. Timoney
P
n−m
and g(z) = ∞
is analytic for |z − a| < δ. Moreover g(a) = am 6= 0, g(z) is
n=m an (z − a)
continuous at z = a and so we can find δ0 > 0, δ0 ≤ δ, with
1
|g(a)| for |z − a| < δ0
2
1
⇒ |g(z)| ≥
|g(a)| for |z − a| < δ0
2
⇒ g(z) 6= 0 for |z − a| < δ0 .
|g(z) − g(a)| ≤
But 0 = f (zn ) and for n large enough (say n > N ) we have |zn − a| < δ0 so that f (zn ) =
(zn − a)m g(zn ) = 0. Thus (for n > N ) zn = 0 or g(zn ) = 0. However, we know g(zn ) 6= 0 and
the zn are distinct so that at most one n can have zn = a. Hence we are faced with a contradiction.
The contradiction arose from assuming that there was any an 6= 0. We must therefore have
an = 0∀n.
(iii) ⇒ (i): Assume now that there is a ∈ G with f (n) (a) = 0 for all n ≥ 0. Then the power
series expansion for f about a (which is valid in a disc D(a, δ) ⊂ G with δ > 0) is
f (z) =
∞
X
f (n) (a)
n=0
n!
(z − a)n = 0 for |z − a| < δ.
Thus f (z) ≡ 0 for |z − a| < δ and differentiating we get f (n) (z) = 0 for n = 0, 1, 2, . . ..
This shows that U = {a ∈ G : f (n) (a) = 0 for all n = 0, 1, 2, . . .} is open (and nonempty).
U is also closed relative to G. To see that take b ∈ G \ U . Then there is some n with f (n) (b) 6= 0.
Now that f (n) is continuous at b and so there is a δ > 0 so that f (n) (z) 6= 0 for all z with
|z − b| < δ. This means none of these z can be in U or in other words D(b, δ) ⊂ G \ U . This
means G \ U is open.
As G is connected, U ⊂ G nonempty and both open and closed relative to G implies U = G.
This means f (n) (z) = 0 for all n ≥ 0 and all z ∈ G. Specifically with n = 0 we have f ≡ 0.
Corollary 3.2 (version with two functions) Let G ⊂ C be open and connected (and nonempty).
Let f, g : G → C be two analytic functions. Then the following are equivalent for f and g:
(i) f ≡ g
(ii) there is an infinite sequence (zn )∞
n=1 of distinct points of G with limn→∞ zn = a ∈ G and
f (zn ) = g(zn )∀n
(iii) there is a point a ∈ G with f (n) (a) = g (n) (a) for n = 0, 1, 2, . . ..
Proof. apply the Identity Theorem 3.1 to the difference f − g.
Remark 3.3 The significance of the Identity Theorem is that an analytic function on a connected
open G ⊂ C is determined on all of G by its behaviour near a single point.
1
Thus if an analytic function is given on one part of G by a formula like f (z) = z−1
and that
formula makes sense and gives an analytic function on a larger connected subset of G then it has
1
to be that f (z) = z−1
also holds in the larger set.
3
Chapter 3 — maximum modulus principle
This is quite different from what happens with continuous functions like f : C → C defined
by
f (z) =
z
z
|z|
|z| < 1
|z| ≥ 1.
Even for C ∞ functions we can have different formulae holding in different places. Consider
g : C → C where

|z| ≤ 1
 0 2 f (z) =
1
|z| > 1.
 exp − |z|−1
The original meaning of the word ‘analytic’ related to this property of analytic functions (one
formula).
Corollary 3.4 If G ⊂ C is a connected open set and f : G → C is analytic and not identically
constant, then the zero set of f
Zf = {z ∈ G : f (z) = 0}
has no accumulation points in G.
Proof. First we should define accumulation point in case you forget it. If S ⊂ C is any set and
a ∈ C, then a is called an accumulation point of S if for each δ > 0
(S \ {a}) ∩ D(a, δ) 6= ∅.
If we a is an accumulation point of S we can choose
z1 ∈ (S \ {a}) ∩ D(a, 1)
1
, |z1 − a|
z2 ∈ (S \ {a}) ∩ D a, min
2
1
z3 ∈ (S \ {a}) ∩ D a, min
, |z2 − a|
3
and (inductively) zn+1 ∈ (S \{a})∩D a, min n1 , |zn − a| . This produces a sequence (zn )∞
n=1
of distinct points zn ∈ S with limn→∞ zn = a. (It is not hard to see that the existence of such a
sequence is equivalent to a being an accumulation point of S.)
Applying this to S = Zf and using Theorem 3.1 we get f ≡ 0.
Corollary 3.5 Let G ⊂ C be open and connected and let K ⊂ G be compact. Let f, g : G → C
be analytic. If the equation f (z) = g(z) has infinitely many solutions z ∈ K, then f ≡ g.
Proof. Choose an infinite sequence (zn )∞
n=1 of distinct points zn ∈ K where f (zn ) = g(zn ).
Since K is compact, the sequence has a convergent subsequence (znj )∞
j=1 with a limit a =
limj→∞ znj ∈ K ⊂ G.
By Corollary 3.2, f ≡ g.
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414 2003–04 R. Timoney
Theorem 3.6 (Maximum modulus theorem, basic version) Let G ⊂ C be a connected open
set and f : G → C analytic. If there is any a ∈ G with |f (a)| ≥ |f (z)| for all z ∈ G, then f is
constant.
Proof. (Another way to state this is that |f (z)| cannot have a maximum in G, unless f is constant.)
Choose δ > 0 so that D(a, δ) ⊂ G. Fix 0 < r < δ and then we have (by the Cauchy integral
formula)
Z
1
f (z)
f (a) =
dz.
2πi |z−a|=r z − a
Write this out in terms of a parametrisation z = a + reiθ with 0 ≤ θ ≤ 2π, dz = ireiθ dθ.
Z 2π
Z 2π
f (a + reiθ )
1
1
dθ
f (a + reiθ ) dθ.
f (a) =
iθ
2πi 0
ire
2π 0
Hence
1
|f (a)| ≤
2π
Z
0
2π
1
|f (a + re )| dθ ≤
2π
iθ
Z
2π
|f (a)| dθ = |f (a)|,
0
using |f (a + reiθ )| ≤ |f (a)|∀θ.
We must therefore have equality in the inequalities. Since the integrand |f (a + reiθ )| is a
continuous function of θ, this implies |f (a + reiθ )| = |f (a)| for all θ.
Put α = Arg(f (a)). Now
|f (a)| = e−iα f (a)
Z
e−iα 2π
=
f (a + reiθ ) dθ
2π 0
Z 2π
1
=
e−iα f (a + reiθ ) dθ
2π 0
Z 2π
1
<|f (a)| = |f (a)| =
<
e−iα f (a + reiθ ) dθ
2π
Z 2π0
1
=
<(e−iα f (a + reiθ )) dθ
2π 0
Z 2π
1
≤
|e−iα f (a + reiθ )| dθ
2π 0
using <w ≤ |w| for w ∈ C
Z 2π
1
≤
|f (a + reiθ )| dθ
2π 0
Z 2π
1
≤
|f (a)| dθ = |f (a)|
2π 0
Thus we must again have equality in all the inequalities and so
<(e−iα f (a + reiθ )) = |e−iα f (a + reiθ )| = |f (a)|
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Chapter 3 — maximum modulus principle
for all θ. Thus =(e−iα f (a + reiθ )) = 0 and e−iα f (a + reiθ ) = |f (a)| or f (a + reiθ ) = eiα |f (a)|.
Thus f (z) is constant for z in the infinite compact subset {z : |z − a| = r} of G. By
Corollary 3.5, it follows that f (z) is constant (on G).
Theorem 3.7 (Maximum modulus theorem, usual version) The absolute value of a nonconstant analytic function on a connected open set G ⊂ C cannot have a local maximum point in
G.
Proof. Let f : G → C be analytic. By a local maximum point for |f | we mean a point a ∈ G
where |f (a)| ≥ |f (z)| holds for all z ∈ D(a, δ) ∩ G, some δ > 0. As G is open, by making
δ > 0 smaller if necessary we can assume D(a, δ) ⊂ G.
By Theorem 3.6, |f (a)| ≥ |f (z)|∀z ∈ D(a, δ) implies f constant on D(a, δ) (since f must
be analytic on D(a, δ) ⊂ G and D(a, δ) is connected open). Then, by the Identity Theorem
(Corollary 3.2), f must be constant.
Corollary 3.8 (Maximum modulus theorem, another usual version) Let G ⊂ C be a bounded
and connected open set. Let f : Ḡ → C be continuous on the closure Ḡ of G and analytic on G.
Then
sup |f (z)| = sup |f (z)|.
z∈∂G
z∈Ḡ
(That is the maximum modulus of the analytic function f (z) is attained on the boundary ∂G.)
Proof. Since G is bounded, its closure Ḡ is closed and bounded, hence compact. |f (z)| is a
continuous real-valued function on the compact set and so supz∈Ḡ |f (z)| < ∞ and the supremum
is attained at some point b ∈ Ḡ. That is |f (b)| ≥ |f (z)|∀z ∈ Ḡ. If b is on the boundary ∂G then
we have
|f (b)| = sup |f (z)| = sup |f (z)|
z∈∂G
z∈Ḡ
but if b ∈ G, then f must be constant by the Identity Theorem 3.1. So in that case supz∈∂G |f (z)| =
supz∈Ḡ |f (z)| is also true.
Theorem 3.9 (Fundamental theorem of algebra) Let p(z) be a nonconstant polynomial (p(z) =
P
n
k
k=0 ak z with an 6= 0 and n ≥ 1). Then the equation p(z) = 0 has a solution z ∈ C.
Proof. Dividing the equation by the coefficient of the highest power of z with a nonzero coefficient (an in the notation above) we can assume without loss of generality that the polynomial is
monic (meaning that the coefficient of the highest power of z is 1) and that the polynomial is
n
p(z) = z +
n−1
X
k=0
ak z k
(n ≥ 1).
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414 2003–04 R. Timoney
Now for |z| = R ≥ R0 = max 1, 2
Pn−1
n
|p(z)| ≥ |z| −
= Rn −
≥ Rn −
|a
|
, we have
k
k=0
n−1
X
|ak ||z|k
k=0
n−1
X
|ak |Rk
k=0
n−1
X
|ak |Rn−1
k=0
using R ≥ 1
n−1
X
n
n−1
= R −R
|ak |
k=0
n
n−1
≥ R − R (R0 /2)
≥ Rn−1 (R − R0 /2) ≥ Rn−1 R/2 = Rn /2 = |z|n /2
Now if p(z) is never zero for z ∈ C, then f (z) = 1/p(z) is an entire function. For |z| ≥ R0
we have
1
1
2
2
|f (z)| =
≤ n = n ≤ n
|p(z)|
|z| /2
|z|
R0
By the maximum modulus theorem (Corollary 3.8)
sup |f (z)| = sup |f (z)| ≤
|z|≤R0
|z|=R0
2
R0n
and so we have |f (z)| ≤ 2/R0n for all z ∈ C. By Liouville’s theorem (1.26) f must be constant.
But then p(z) = 1/f (z) is also constant, contradicting our hypotheses.
The contradiction arose by assuming p(z) was never 0. So p(z) = 0 for some z ∈ C.
Remark 3.10 There are many ways to prove the fundamental theorem, but all (or at least most)
rely on the same first step as above — for |z| large, p(z) behaves like its term of highest degree
an z n . To make things a little simpler, we took an = 1 but that is not really essential. The idea is
that, when |z| is large the highest term outweighs the combination of the lower degree terms.
Also, although we used the maximum modulus theorem to get the same bound for f (z) for
|z| ≤ R0 as for |z| ≥ R0 , we could have just used compactness to get sup|z|≤R0 |f (z)| < ∞. (We
used an argument like that once before in the proof of the winding number version of Cauchy’s
theorem (1.30 claim 2 in the proof).
P
Corollary 3.11 If p(z) = nk=0 ak z k is a polynomial of degree n (meaning an 6= 0) then p(z)
can be factored
n
Y
p(z) = an (z − αj )
j=1
for some α1 , α2 , . . . , αn ∈ C.
Chapter 3 — maximum modulus principle
7
Proof. For n = 0 the result is true (provided we interpret the empty product as 1).
By the Fundamental Theorem of algebra 3.9, there is some α1 ∈ C with p(α1 ) = 0. Then it
follows from the remainder theorem that z −α1 divides p(z). In other words p(z) = (z −α1 )q(z)
where q(z) has degree n − 1 and leading coefficient an (same as for p(z)).
If we arrange the proof more formally as induction on the degree, we have
Qnthe starting n = 0
and the induction step. Applying the result from degree n − 1 to q(z) = an j=2 (z − αj ) we are
done.
Richard M. Timoney
December 3, 2003