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Chapter II The Expansion of Functions and Picard Theorems
CHAPTER II
THE EXPANSION OF FUNCTIONS AND PICARD THEOREMS
As a preamble to the expansion of a meromorphic and an entire function,
we recall the following. 2.1 RESIDUES
Let f(z) have a pole of order k at z = z,. Write
and
Thus
dk-1 [(z - z,)kf(z)]= (k - l)!a.-l dzk-l (k I)! al(z - zo)2 ... ,
+
+
+
2!
dk-1
26
+ - k!al !o ( z
- z,)
2.2-2.3 Expansion of a Meromorphic Function
27
and
2.2
EXPANSION OF A MEROMORPHIC FUNCTION
Since a meromorphic function is analytic in a region except possibly at a finite number of poles, we have the following theorem. 2.2.1 Theorem. Let f ( z ) be a function whose only singularities, except at
infinity, are poles. Suppose all poles are simple. Let them be a,, a,, . . . and be ordered such that 0 < I a, I 5 I a, I 5 and let the residues at the poles be 6 , , b,, . . . . Suppose there is a sequence of contours C,,, such that C,, includes a, , a,, . . . , a,, , but no other poles. Let the minimum distance Rn of C, from the origin approach 00 with n, while Ln the length of C,>is O ( R , ) and such that on C , , ,f ( z ) = o(R,). This last condition is satisfied if, for example, f ( z ) is bounded on all C,, Then
-
.
for all z except poles. Proof. Consider
I=
1
___
2ni
f ( w ) dw c, w ( w - z )
where z E C,,. The integrand has poles at arn,0, and w bmlam(a, - z ) , - f(O)/z,f ( z ) l z , respectively
==
z, with residues
and
by the conditions imposed. Thus
Note, that the series converges uniformly inside any closed contour such that all poles are inside. [7
11.' The Expansion of Functions and Picard Theorems
28
2.3 POLES AND ZEROS OF A MEROMORPHIC FUNCTION
2.3.1 Theorem. If f ( z ) is analytic in and on a closed contour C apart from a finite number of poles and if f(z) # 0, z E C, then
where N is the number of zeros in C (zero of order m counted m times) and P is the number of poles in C (pole of order m counted m times).
Proof. Let z
=a
be a zero of order m. Then in the neighborhood of z f ( z >= ( z
- a>"&>
and analytic in the neighborhood of z
= a,
for g ( z ) # 0 = a.
Thus
The last term is analytic at z = a, therefore f ' ( z ) / f ( z )has a simple pole at z = a, with residue m. The sum of the residues at the zeros of f(z) is N (the number of zeros). Similarly by writing -m for m, the sum of the residues at the poles of f ( z ) is -P (number of poles). Iff(z) is an entire function, the number of zeros
2.3.2. Corollary. If # ( z ) is analytic in and on C and if f ( z ) has zeros at a l , . . . , a p and poles at b,, . . . , b,, multiplicity being counted as before,
for if z z
=
=a
is a zero of order m, we have in the neighborhood of - a)mg(z) where g ( z ) # 0 and analytic. Thus
a, f ( z ) = ( z
The last term is analytic at z = a, therefore the left-hand side has a simple pole at z = a with residue m$(a). Applying the previous theorem we have the result.
29
2.4 Expansion of an Entire Function
2.4 EXPANSION OF AN ENTIRE FUNCTION AS AN INFINITE PRODUCT
Supposef(z) has simple zeros at the points a,, a 2 , . . . , a,. In the neighborhood of a , , f(z) = (z - a,)g(z), where g(z) is analytic and nonzero. Thus f‘(z> f(z)
-
z
1 - an
+-g‘g(z)(z)
with g’(z)/g(z) being analytic at a,, . Hence f’(z)/f(z) has a simple pole at z = a, with residue 1. Suppose f’(z)/f(z) is a function of the type considered in the expansion of a meromorphic function f(z). Then P(z) = f’(z)/f(z) has poles at a , , a 2 , . . . , a, and
Clearly P(0) = f’(O)/f(O) and b,, = 1 for all n. Thus
Integrating from 0 to z along a path not passing through a pole, we have logf(z) - logf(0)
=
+ f {log(z -
zf’(o) f(0)
n=l
where the value of the logarithms depend upon the path. Taking exponentials,
As an example, consider f(z)
sin z
= -Z
fi’ (1 -$)expnn,
n=--m
or sinz=z
Z
m
n-1
(1
-=). Z2
The next theorem called Rouchk‘s theorem follows somewhat naturally.
30
11. The Expansion of Functions and Picard Theorems
2.5 Theorem. Iff(z) and g(z) are analytic in and on a closed contour C, and if I g(z) I < If(z) 1 on C, then f(z) and f(z) g(z) have the same number of zeros inside C.
+
Proof. Let 4 ( z ) = g(z)/f(z). Then I +(z) I < 1 on C. Note that neither f(z) norf(z) g(z) has a zero on C. If the number of zeros off(z) g(z) inside C is denoted by N ‘ , then
+
+
=N+-
2ni
4’{1 - 4
+
+2
- . . . } dz.
Also,
thus by uniform convergence of the series we have N ‘
=
N.
0
The following theorem is now a consequence.
2.6 A Theorem of Hurwitz. Let f;,(z) be a sequence of functions analytic in a region D bounded by a simple closed contour. Let fn(z) tend to f(z) uniformly. Assume f(z) $0, and let zo be an interior point of D.Then z, is a zero off(z) if and only if z, is a limit point of zeros ofj;(z), n = 1, 2, . . . and points which are zeros for infinite n are counted as limit points. Proof. Choose e small and such that the circle I z - zo I = e is in D and contains or has on it no zero off(z) except possibly z,. Then If(z) I has a positive lower bound on the circle, i.e., If@) I L m > 0. Having fixed
e and m, choose N
If,(z) -f(z)
I
so large that
for n L N , z E { I z
Since
f ( z ) =f(z)
+ {fn(z> -f(z>),
-
zo I
=
e}.
31
2.7 Picard Theorems for Functions of Finite Order
Jl(z) has the same number of zeros in the circle asf(z). This follows from Rouchts theorem because f(z) and f,,(z) -f(z) are analytic in and on a closed contour C in a region D and If,(z) -f(z) I < If(z) I. Thus if f(zO) = 0, then f,,(z) has exactly one zero in C for all n 2 N so that z0 is a limit point of zeros off,(z). Also, iff(z,) # 0, thenf,(z,,) # 0 in C. 0
2.7 The next group of theorems although elementary in nature, examine more carefully the relation between order and the number of roots of an equation of the form f(z) = AP(z) where f(z) is entire, A is a constant, and P(z) is a nonzero polynomial. The only deficiency in the theorems is that they do not deal with functions of infinite order, a ramification which will be dealt with later on in the survey. 2.7.1 Picard Theorems. It will be useful to establish first of all, a few lemmas concerning inequalities for coefficients of power series. . . . anzn . . Write Consider the entire functionf(z) = a, a,z
+ + a,,= an + iBn
and
z
= r(cos
Then f(z)
m
=
C
n=o
(orn
6
+ i sin 0),
f(z> = U(r, 0) U(r, 0)
+C m
= a0
+ -
2 0,
+ i,Bn)rn(cos8 + i sin 0 ) n .
If we write then
r
+
n=1
+ Wr, 0
(or, cos n0 - B,, sin nO)rn.
(1)
Fixing r and integrating with respect to 0 from 0 to 2n, we have Llg =
1 J 2 n U(r, 0) do. 2n
0
To compute orpr p 2 1 multiply ( I ) by cosp0 and integrate with respect to 0 from 0 to 2n. Thus U(r, 0) cos p6 d8 = aprp
32
11. The Expansion of Functions and Picard Theorems
from which
=-j nrp
a
and similarly Pp
=-
nrp
Also,
2n
V ( r , e) COspe dB
p = 1,2, ..
j2nV(r,0 ) sinpBd0,
p
=
1, 2, . . .
.
12n
2a0 fuprp = U ( r , e)(i f cospe) dB n o
and 2a0
j
Bprp = - 2n V ( r , e ) ( l f sinp0) do. n o
Note that the factors multiplying U(r, 0 ) are now nonnegative. Let p ( r ) = maxosB52nU(r, 0 ) on the circle of radius r. Then
j2’ f cos pe) d0
2u, f uprp 5 ’(‘) (1 n o
= 2p (r).
2.7.2 Lemma. If U(r, 0 ) the real part of an entire function f ( z ) satisfies U(r, 0 ) 5 p ( r ) 5 Crs,
S > 0,
for all r > N , then f(z) is a polynomial of degree not exceeding n = [S], i.e., the integer part of 6.
Proof. Since
I up I
1% I 5
and
I /-II,5 2 { p ( r ) - uo}/rP,we
2 (Cr6 - uo) ,p
and
IBPI
+
5
have that
2(Crs - uo) rp
If p > [a], since p is an integer, p 2 [S] 1 > 6 hence u p , r - + o o , and up iBP = ap = 0 for p > [d] = n. 0
+
PP
-
0,
33
2.7 Picard Theorems for Functions of Finite Order
2.7.3 Lemma. I f f ( z ) is a transcendental entire function and P ( z ) and Q ( z )
+
are polynomials of degree m, n, respectively and if P ( z ) 0, then the order el of P ( z ) f ( z ) Q ( z ) coincides with the order e off(z), i.e.,
+
el = e. Proof. Writing
M ( r ) = max I f ( z ) I2153
we confine ourselves to values of the functions at points on the circle I z I = r. Let a,z*, b,zn denote, respectively, the leading terms in the polynomials P(z) and Q(z). Taking E = & in the inequalities for polynomials previously evaluated, we assert for I z I = r > r o , that
However, at z1 (on the same circle) at which 1 P(z)f(z) its maximum M l ( r ) , we have Mi(r)
=
+ Q(z) I attains
I P(zilf(zi) + Q(zi) I I 8 I a, I rrnM(r)+ 8 I bn I rn*
Thus
Sincef(z) is a transcendental entire function it has been proved that M ( r ) increases faster than the maximum modulus of any polynomial and hence faster than any power of r. Therefore each { . } + 1 as r 00 and hence < 2 and the left hand for r sufficiently large, the right-hand { } > Q , i.e.,
-
{ - . a }
i I a, I rmM(r)L MI@)5 3 I a, I rnM(r).
34
11. The Expansion of Functions and Picard Theorems
We now require
- log log M(r) lim r+m log r
Y
i.e.,
and each {. . . } -+ 1 as r + 00. Hence the right-hand { left-hand { } > 4, and
---
9
.} < 2
and the
Q log M ( r ) 5 log M,(r) I 2 log M(r). Taking logarithms again,
- log log M(r) lim
,-+m
consequently [P(z)fOl.
logr
log log M l ( r ) 5lim logr
,+m
e = el and
~
lim ,-+-
log log M ( r ) logr '
the orders off(z) and P(z)f(z)
0
+ Q(z) are equal
2.7.4 Lemma. The order of an entire function
where P ( z ) , Q(z), and g(z) are polynomials and P(z) $0, is equal to the degree of g(z). Proof. From the previous lemma, the order der of $(.) = e@). We need to show that
- log log Ml(r) lim
r+.n
where max,,,,,
and
logr
e off(z)
coincides with the or-
=n
1 $(z) I = M,(r) and n is the degree of g(z). Set
z
= r(cos
0
+ i sin 0).
35
2.7 Picard Theorems for Functions of Finite Order
By hypothesis, en = 1 c, g(z) = =
I # 0,
thus
goekrk(cos n
ak
n Z O
+ i sin Cr,)(cos kO + i sin kO) + ke) -/-
Qkrk(COS(ai,
i S i l l ( a k $-
k0))
and
and
We require now log rnaxlzI,,I $(z)
hence for
E
I.
For fixed r and 0 5 O 5 2n
in (0, 1) and r > r(E),
and Let zo be a point on I z I are n such points). Then 1% Mi(r) 2 1%
=
r such that cos(a,,
I d(zo) I = @ J n
+
2 enrn
-
n-1
@krk cos(ffk
zo
Z O
+ no) = 1 (actually there
+ ke).
n-1
@krk
1 + -
en
> enrn(l - E ) Thus
ep(l
- E)
and -
lirn
r+m
...
r for r > r(E).
< log M l ( r ) < enrn(l + E )
log log M , ( r ) logr
= n.
0
en
11. The Expansion of Functions and Picard Theorems
36
2.7.5 Lemma. If g(z) is an entire function and if the order of the function f(z) = eg(z)is finite, then g(z) is a polynomial and hence the order off(z) is an integer.
Pvoof. If z
=
r (cos 0
+ i sin 0) and
1% I f(z) I Let
max If(z) IzI-r
I = M(r)
Then
U(r, 0) is the real part of g ( z ) , then =
and
w, 0). max
osean
U(r, 0)
= p(r).
log M ( r ) = p ( r ) .
Suppose 6 is the order off(z). Then - log
lim
r+m
Given
E
log M ( r ) log r
= 6.
> 0, 34.9) > 1 such that for r > r(E) log log M ( r ) log r
i.e.,
< 8 + E
log log M ( r ) < log(r6+")
or
log M ( r ) < rd+E.
Thus from (I),
p ( r ) < r6+"
for r > r(E).
From Lemma 2.7.2 it follows that g ( z ) is a polynomial the degree n of which satisfies n 5 [S E ] and E is arbitrarily small. Thus n 5 [6] and (by Lemma 2.7.4) the order of f ( z ) coincides with n, i.e., 6 is integral. 0
+
2.7.6 Theorem. Let f ( z ) denote a transcendental entire function, the order 6 of which is finite but nonintegral. Then if P(z) is a nonzero polynomial, the equation f(z)=
has infinitely many roots for every complex number A (no exceptions).
Proof. Suppose 3 A
=
A, for which f ( z ) = A,P(z) has only finitely many
37
2.7 Picard Theorems for Functions of Finite Order
roots if any at all. Then the entire functionf(z) - A,P(z) has only finitely many zeros. Therefore by a previous theorem we may write f(z) - A,P(z)
=
Q(z)eg(z)
where Q(z) is a nonzero polynomial (equal to I iff - AoP has no zeros). Further g(z) is an entire function, and we have f(z)
=
A,P(z) -t Q(z)eg'.'.
Thus the order 6 off(z) coincides with the order of eg(z)which, being finite by a previous lemma, is integral. The contradiction proves the theorem. 0 2.7.7 Theorem. Iff(z) is a transcendental entire function of finite integral order n and if P(z) is a nonzero polynomial, then
f(z) = AP(z) has infinitely many roots for every complex number A , with the possible exception of one value.
Proof. Suppose there exists at least two values a, b at whichf(z) = A P ( z ) has only finitely many roots. Then f(z) - aP(z) and f(z) - bP(z) have only finitely many zeros. Thus and
f(z) - aP(z)
=
Q,(z)e"'z)
(1 1
f(z) - bP(z)
=
Qz(z)eez(z)
(2)
where Ql(z) and Qz(z) are nonzero polynomials and gl(z) and gz(z) are entire functions. Hence the orders of egi(.) and egz(z)coincide with the order n off(z). We also conclude that g,(z) and g,(z) are polynomials of degree n. Hence n 2 1 since for n = 0, g,(z) and gz(z) would be constants and f(z) would be a polynomial but not a transcendental entire function. Subtracting (1) from (2), Q1(z)eQ1(')- Qz(z)eQa(-') = (b - a)P(z)
=
R(z),
where R(z) is a nonzero polynomial since b # a and by hypothesis P(z) # 0. We need to show that this equation is not possible for Q,(z), Q,(z) and R ( z ) nonzero polynomials and gl(z), gz(z) polynomials of degree greater than or equal to 1. Differentiating (Qi'
+ Qigi')e@~ (Qz' + Qzgz')eQa -
=
R'
38
11. The Expansion of Functions and Picard Theorems
also =
Ql egl - Q ,
R.
Regarding these as a system of two equations in the unknowns eel and the determinant of the system d ( z ) is given by Qi(Qz’ = -QzQi’ =
+ Qgz‘)
+ QiQz‘
- QdQi‘
eg:,
+ Qigi’)
- QiQzki’
- gz’).
We must now show that d ( z ) $0. If we assume that A ( z ) = 0, divide by -Q,Q, to obtain el’
Qz’
Qz
Qi
+ (8,
- g,)‘
= 0.
Integrating this equation we have Qi log -
Q,
+ g , - g, = constant = C,
Thus
However, by dividing Eqs. ( I ) and (2) we obtain
Q @1-9n 1
=
Q2
Thus
f-aP f-bP
--
and
f-aP f-bP‘
~
-c
f ( z ) ( l - C ) = P(z)(a - bC).
Since b # a , we have C # 1 and hence
which contradicts the hypothesis, forf(z) is a transcendental entire function and P ( z ) is a polynomial. Thus d ( z ) $ 0. Solving for egl and eh, we obtain
2.7 Picard Theorems for Functions of Finite Order
39
However, these equations contain a contradiction since the left-hand side is a transcendental entire function and the function on the right-hand side is a rational function, and thus a polynomial since it is entire. We can say [for P(z) = I ] thatf(z) takes on all values an infinite number of times with one possible exception. 0 The previous two theorems are special cases of the following theorem. 2.7.8 Theorem. Let Q(z) be a transcendental meromorphic function. For every complex number A (finite or infinite) with two possible exceptions, Q(z) = A has infinitely many roots.
Proof. If Q ( z ) is an entire function, it does not become infinite at any z. Thus A = 00 is an exceptional value for every entire function. Consequently, on the basis of the previous theorem there may exist at most one finite exceptional value. If f(z) is a transcendental entire function and P(z) is a nonzero polynomial, the meromorphic function Q(z) = f(z)/P(z)becomes infinite only at a finite number of points, namely the zeros of P(z). Thus 00 is an exceptional value, and further, one other finite exceptional value is possible. Thus f ( z ) / P ( z )= A or f(z) = A P ( z ) has infinitely many roots for every A except possibly a single finite value of A . Letf(z) and g(z) denote two transcendental entire functions whose ratio is not rational. Hence f(z)/g(z) is meromorphic. Thus f(z)/g(z) = A has infinitely many roots with two possible exceptions. 0
THE EXPANSION OF FUNCTIONS AND PICARD THEOREMS
As a preamble to the expansion of a meromorphic and an entire function,
we recall the following. 2.1 RESIDUES
Let f(z) have a pole of order k at z = z,. Write
and
Thus
dk-1 [(z - z,)kf(z)]= (k - l)!a.-l dzk-l (k I)! al(z - zo)2 ... ,
+
+
+
2!
dk-1
26
+ - k!al !o ( z
- z,)
2.2-2.3 Expansion of a Meromorphic Function
27
and
2.2
EXPANSION OF A MEROMORPHIC FUNCTION
Since a meromorphic function is analytic in a region except possibly at a finite number of poles, we have the following theorem. 2.2.1 Theorem. Let f ( z ) be a function whose only singularities, except at
infinity, are poles. Suppose all poles are simple. Let them be a,, a,, . . . and be ordered such that 0 < I a, I 5 I a, I 5 and let the residues at the poles be 6 , , b,, . . . . Suppose there is a sequence of contours C,,, such that C,, includes a, , a,, . . . , a,, , but no other poles. Let the minimum distance Rn of C, from the origin approach 00 with n, while Ln the length of C,>is O ( R , ) and such that on C , , ,f ( z ) = o(R,). This last condition is satisfied if, for example, f ( z ) is bounded on all C,, Then
-
.
for all z except poles. Proof. Consider
I=
1
___
2ni
f ( w ) dw c, w ( w - z )
where z E C,,. The integrand has poles at arn,0, and w bmlam(a, - z ) , - f(O)/z,f ( z ) l z , respectively
==
z, with residues
and
by the conditions imposed. Thus
Note, that the series converges uniformly inside any closed contour such that all poles are inside. [7
11.' The Expansion of Functions and Picard Theorems
28
2.3 POLES AND ZEROS OF A MEROMORPHIC FUNCTION
2.3.1 Theorem. If f ( z ) is analytic in and on a closed contour C apart from a finite number of poles and if f(z) # 0, z E C, then
where N is the number of zeros in C (zero of order m counted m times) and P is the number of poles in C (pole of order m counted m times).
Proof. Let z
=a
be a zero of order m. Then in the neighborhood of z f ( z >= ( z
- a>"&>
and analytic in the neighborhood of z
= a,
for g ( z ) # 0 = a.
Thus
The last term is analytic at z = a, therefore f ' ( z ) / f ( z )has a simple pole at z = a, with residue m. The sum of the residues at the zeros of f(z) is N (the number of zeros). Similarly by writing -m for m, the sum of the residues at the poles of f ( z ) is -P (number of poles). Iff(z) is an entire function, the number of zeros
2.3.2. Corollary. If # ( z ) is analytic in and on C and if f ( z ) has zeros at a l , . . . , a p and poles at b,, . . . , b,, multiplicity being counted as before,
for if z z
=
=a
is a zero of order m, we have in the neighborhood of - a)mg(z) where g ( z ) # 0 and analytic. Thus
a, f ( z ) = ( z
The last term is analytic at z = a, therefore the left-hand side has a simple pole at z = a with residue m$(a). Applying the previous theorem we have the result.
29
2.4 Expansion of an Entire Function
2.4 EXPANSION OF AN ENTIRE FUNCTION AS AN INFINITE PRODUCT
Supposef(z) has simple zeros at the points a,, a 2 , . . . , a,. In the neighborhood of a , , f(z) = (z - a,)g(z), where g(z) is analytic and nonzero. Thus f‘(z> f(z)
-
z
1 - an
+-g‘g(z)(z)
with g’(z)/g(z) being analytic at a,, . Hence f’(z)/f(z) has a simple pole at z = a, with residue 1. Suppose f’(z)/f(z) is a function of the type considered in the expansion of a meromorphic function f(z). Then P(z) = f’(z)/f(z) has poles at a , , a 2 , . . . , a, and
Clearly P(0) = f’(O)/f(O) and b,, = 1 for all n. Thus
Integrating from 0 to z along a path not passing through a pole, we have logf(z) - logf(0)
=
+ f {log(z -
zf’(o) f(0)
n=l
where the value of the logarithms depend upon the path. Taking exponentials,
As an example, consider f(z)
sin z
= -Z
fi’ (1 -$)expnn,
n=--m
or sinz=z
Z
m
n-1
(1
-=). Z2
The next theorem called Rouchk‘s theorem follows somewhat naturally.
30
11. The Expansion of Functions and Picard Theorems
2.5 Theorem. Iff(z) and g(z) are analytic in and on a closed contour C, and if I g(z) I < If(z) 1 on C, then f(z) and f(z) g(z) have the same number of zeros inside C.
+
Proof. Let 4 ( z ) = g(z)/f(z). Then I +(z) I < 1 on C. Note that neither f(z) norf(z) g(z) has a zero on C. If the number of zeros off(z) g(z) inside C is denoted by N ‘ , then
+
+
=N+-
2ni
4’{1 - 4
+
+2
- . . . } dz.
Also,
thus by uniform convergence of the series we have N ‘
=
N.
0
The following theorem is now a consequence.
2.6 A Theorem of Hurwitz. Let f;,(z) be a sequence of functions analytic in a region D bounded by a simple closed contour. Let fn(z) tend to f(z) uniformly. Assume f(z) $0, and let zo be an interior point of D.Then z, is a zero off(z) if and only if z, is a limit point of zeros ofj;(z), n = 1, 2, . . . and points which are zeros for infinite n are counted as limit points. Proof. Choose e small and such that the circle I z - zo I = e is in D and contains or has on it no zero off(z) except possibly z,. Then If(z) I has a positive lower bound on the circle, i.e., If@) I L m > 0. Having fixed
e and m, choose N
If,(z) -f(z)
I
so large that
for n L N , z E { I z
Since
f ( z ) =f(z)
+ {fn(z> -f(z>),
-
zo I
=
e}.
31
2.7 Picard Theorems for Functions of Finite Order
Jl(z) has the same number of zeros in the circle asf(z). This follows from Rouchts theorem because f(z) and f,,(z) -f(z) are analytic in and on a closed contour C in a region D and If,(z) -f(z) I < If(z) I. Thus if f(zO) = 0, then f,,(z) has exactly one zero in C for all n 2 N so that z0 is a limit point of zeros off,(z). Also, iff(z,) # 0, thenf,(z,,) # 0 in C. 0
2.7 The next group of theorems although elementary in nature, examine more carefully the relation between order and the number of roots of an equation of the form f(z) = AP(z) where f(z) is entire, A is a constant, and P(z) is a nonzero polynomial. The only deficiency in the theorems is that they do not deal with functions of infinite order, a ramification which will be dealt with later on in the survey. 2.7.1 Picard Theorems. It will be useful to establish first of all, a few lemmas concerning inequalities for coefficients of power series. . . . anzn . . Write Consider the entire functionf(z) = a, a,z
+ + a,,= an + iBn
and
z
= r(cos
Then f(z)
m
=
C
n=o
(orn
6
+ i sin 0),
f(z> = U(r, 0) U(r, 0)
+C m
= a0
+ -
2 0,
+ i,Bn)rn(cos8 + i sin 0 ) n .
If we write then
r
+
n=1
+ Wr, 0
(or, cos n0 - B,, sin nO)rn.
(1)
Fixing r and integrating with respect to 0 from 0 to 2n, we have Llg =
1 J 2 n U(r, 0) do. 2n
0
To compute orpr p 2 1 multiply ( I ) by cosp0 and integrate with respect to 0 from 0 to 2n. Thus U(r, 0) cos p6 d8 = aprp
32
11. The Expansion of Functions and Picard Theorems
from which
=-j nrp
a
and similarly Pp
=-
nrp
Also,
2n
V ( r , e) COspe dB
p = 1,2, ..
j2nV(r,0 ) sinpBd0,
p
=
1, 2, . . .
.
12n
2a0 fuprp = U ( r , e)(i f cospe) dB n o
and 2a0
j
Bprp = - 2n V ( r , e ) ( l f sinp0) do. n o
Note that the factors multiplying U(r, 0 ) are now nonnegative. Let p ( r ) = maxosB52nU(r, 0 ) on the circle of radius r. Then
j2’ f cos pe) d0
2u, f uprp 5 ’(‘) (1 n o
= 2p (r).
2.7.2 Lemma. If U(r, 0 ) the real part of an entire function f ( z ) satisfies U(r, 0 ) 5 p ( r ) 5 Crs,
S > 0,
for all r > N , then f(z) is a polynomial of degree not exceeding n = [S], i.e., the integer part of 6.
Proof. Since
I up I
1% I 5
and
I /-II,5 2 { p ( r ) - uo}/rP,we
2 (Cr6 - uo) ,p
and
IBPI
+
5
have that
2(Crs - uo) rp
If p > [a], since p is an integer, p 2 [S] 1 > 6 hence u p , r - + o o , and up iBP = ap = 0 for p > [d] = n. 0
+
PP
-
0,
33
2.7 Picard Theorems for Functions of Finite Order
2.7.3 Lemma. I f f ( z ) is a transcendental entire function and P ( z ) and Q ( z )
+
are polynomials of degree m, n, respectively and if P ( z ) 0, then the order el of P ( z ) f ( z ) Q ( z ) coincides with the order e off(z), i.e.,
+
el = e. Proof. Writing
M ( r ) = max I f ( z ) I2153
we confine ourselves to values of the functions at points on the circle I z I = r. Let a,z*, b,zn denote, respectively, the leading terms in the polynomials P(z) and Q(z). Taking E = & in the inequalities for polynomials previously evaluated, we assert for I z I = r > r o , that
However, at z1 (on the same circle) at which 1 P(z)f(z) its maximum M l ( r ) , we have Mi(r)
=
+ Q(z) I attains
I P(zilf(zi) + Q(zi) I I 8 I a, I rrnM(r)+ 8 I bn I rn*
Thus
Sincef(z) is a transcendental entire function it has been proved that M ( r ) increases faster than the maximum modulus of any polynomial and hence faster than any power of r. Therefore each { . } + 1 as r 00 and hence < 2 and the left hand for r sufficiently large, the right-hand { } > Q , i.e.,
-
{ - . a }
i I a, I rmM(r)L MI@)5 3 I a, I rnM(r).
34
11. The Expansion of Functions and Picard Theorems
We now require
- log log M(r) lim r+m log r
Y
i.e.,
and each {. . . } -+ 1 as r + 00. Hence the right-hand { left-hand { } > 4, and
---
9
.} < 2
and the
Q log M ( r ) 5 log M,(r) I 2 log M(r). Taking logarithms again,
- log log M(r) lim
,-+m
consequently [P(z)fOl.
logr
log log M l ( r ) 5lim logr
,+m
e = el and
~
lim ,-+-
log log M ( r ) logr '
the orders off(z) and P(z)f(z)
0
+ Q(z) are equal
2.7.4 Lemma. The order of an entire function
where P ( z ) , Q(z), and g(z) are polynomials and P(z) $0, is equal to the degree of g(z). Proof. From the previous lemma, the order der of $(.) = e@). We need to show that
- log log Ml(r) lim
r+.n
where max,,,,,
and
logr
e off(z)
coincides with the or-
=n
1 $(z) I = M,(r) and n is the degree of g(z). Set
z
= r(cos
0
+ i sin 0).
35
2.7 Picard Theorems for Functions of Finite Order
By hypothesis, en = 1 c, g(z) = =
I # 0,
thus
goekrk(cos n
ak
n Z O
+ i sin Cr,)(cos kO + i sin kO) + ke) -/-
Qkrk(COS(ai,
i S i l l ( a k $-
k0))
and
and
We require now log rnaxlzI,,I $(z)
hence for
E
I.
For fixed r and 0 5 O 5 2n
in (0, 1) and r > r(E),
and Let zo be a point on I z I are n such points). Then 1% Mi(r) 2 1%
=
r such that cos(a,,
I d(zo) I = @ J n
+
2 enrn
-
n-1
@krk cos(ffk
zo
Z O
+ no) = 1 (actually there
+ ke).
n-1
@krk
1 + -
en
> enrn(l - E ) Thus
ep(l
- E)
and -
lirn
r+m
...
r for r > r(E).
< log M l ( r ) < enrn(l + E )
log log M , ( r ) logr
= n.
0
en
11. The Expansion of Functions and Picard Theorems
36
2.7.5 Lemma. If g(z) is an entire function and if the order of the function f(z) = eg(z)is finite, then g(z) is a polynomial and hence the order off(z) is an integer.
Pvoof. If z
=
r (cos 0
+ i sin 0) and
1% I f(z) I Let
max If(z) IzI-r
I = M(r)
Then
U(r, 0) is the real part of g ( z ) , then =
and
w, 0). max
osean
U(r, 0)
= p(r).
log M ( r ) = p ( r ) .
Suppose 6 is the order off(z). Then - log
lim
r+m
Given
E
log M ( r ) log r
= 6.
> 0, 34.9) > 1 such that for r > r(E) log log M ( r ) log r
i.e.,
< 8 + E
log log M ( r ) < log(r6+")
or
log M ( r ) < rd+E.
Thus from (I),
p ( r ) < r6+"
for r > r(E).
From Lemma 2.7.2 it follows that g ( z ) is a polynomial the degree n of which satisfies n 5 [S E ] and E is arbitrarily small. Thus n 5 [6] and (by Lemma 2.7.4) the order of f ( z ) coincides with n, i.e., 6 is integral. 0
+
2.7.6 Theorem. Let f ( z ) denote a transcendental entire function, the order 6 of which is finite but nonintegral. Then if P(z) is a nonzero polynomial, the equation f(z)=
has infinitely many roots for every complex number A (no exceptions).
Proof. Suppose 3 A
=
A, for which f ( z ) = A,P(z) has only finitely many
37
2.7 Picard Theorems for Functions of Finite Order
roots if any at all. Then the entire functionf(z) - A,P(z) has only finitely many zeros. Therefore by a previous theorem we may write f(z) - A,P(z)
=
Q(z)eg(z)
where Q(z) is a nonzero polynomial (equal to I iff - AoP has no zeros). Further g(z) is an entire function, and we have f(z)
=
A,P(z) -t Q(z)eg'.'.
Thus the order 6 off(z) coincides with the order of eg(z)which, being finite by a previous lemma, is integral. The contradiction proves the theorem. 0 2.7.7 Theorem. Iff(z) is a transcendental entire function of finite integral order n and if P(z) is a nonzero polynomial, then
f(z) = AP(z) has infinitely many roots for every complex number A , with the possible exception of one value.
Proof. Suppose there exists at least two values a, b at whichf(z) = A P ( z ) has only finitely many roots. Then f(z) - aP(z) and f(z) - bP(z) have only finitely many zeros. Thus and
f(z) - aP(z)
=
Q,(z)e"'z)
(1 1
f(z) - bP(z)
=
Qz(z)eez(z)
(2)
where Ql(z) and Qz(z) are nonzero polynomials and gl(z) and gz(z) are entire functions. Hence the orders of egi(.) and egz(z)coincide with the order n off(z). We also conclude that g,(z) and g,(z) are polynomials of degree n. Hence n 2 1 since for n = 0, g,(z) and gz(z) would be constants and f(z) would be a polynomial but not a transcendental entire function. Subtracting (1) from (2), Q1(z)eQ1(')- Qz(z)eQa(-') = (b - a)P(z)
=
R(z),
where R(z) is a nonzero polynomial since b # a and by hypothesis P(z) # 0. We need to show that this equation is not possible for Q,(z), Q,(z) and R ( z ) nonzero polynomials and gl(z), gz(z) polynomials of degree greater than or equal to 1. Differentiating (Qi'
+ Qigi')e@~ (Qz' + Qzgz')eQa -
=
R'
38
11. The Expansion of Functions and Picard Theorems
also =
Ql egl - Q ,
R.
Regarding these as a system of two equations in the unknowns eel and the determinant of the system d ( z ) is given by Qi(Qz’ = -QzQi’ =
+ Qgz‘)
+ QiQz‘
- QdQi‘
eg:,
+ Qigi’)
- QiQzki’
- gz’).
We must now show that d ( z ) $0. If we assume that A ( z ) = 0, divide by -Q,Q, to obtain el’
Qz’
Qz
Qi
+ (8,
- g,)‘
= 0.
Integrating this equation we have Qi log -
Q,
+ g , - g, = constant = C,
Thus
However, by dividing Eqs. ( I ) and (2) we obtain
Q @1-9n 1
=
Q2
Thus
f-aP f-bP
--
and
f-aP f-bP‘
~
-c
f ( z ) ( l - C ) = P(z)(a - bC).
Since b # a , we have C # 1 and hence
which contradicts the hypothesis, forf(z) is a transcendental entire function and P ( z ) is a polynomial. Thus d ( z ) $ 0. Solving for egl and eh, we obtain
2.7 Picard Theorems for Functions of Finite Order
39
However, these equations contain a contradiction since the left-hand side is a transcendental entire function and the function on the right-hand side is a rational function, and thus a polynomial since it is entire. We can say [for P(z) = I ] thatf(z) takes on all values an infinite number of times with one possible exception. 0 The previous two theorems are special cases of the following theorem. 2.7.8 Theorem. Let Q(z) be a transcendental meromorphic function. For every complex number A (finite or infinite) with two possible exceptions, Q(z) = A has infinitely many roots.
Proof. If Q ( z ) is an entire function, it does not become infinite at any z. Thus A = 00 is an exceptional value for every entire function. Consequently, on the basis of the previous theorem there may exist at most one finite exceptional value. If f(z) is a transcendental entire function and P(z) is a nonzero polynomial, the meromorphic function Q(z) = f(z)/P(z)becomes infinite only at a finite number of points, namely the zeros of P(z). Thus 00 is an exceptional value, and further, one other finite exceptional value is possible. Thus f ( z ) / P ( z )= A or f(z) = A P ( z ) has infinitely many roots for every A except possibly a single finite value of A . Letf(z) and g(z) denote two transcendental entire functions whose ratio is not rational. Hence f(z)/g(z) is meromorphic. Thus f(z)/g(z) = A has infinitely many roots with two possible exceptions. 0