COMPLEX FUNCTIONS
33
4. What is the genera.1form of a rational funct,ion which has absolute
value 1 on the circle lzl = l ? I n particular, how are the zeros and poles
related to each other?
5. If a rational function is real on lzl = 1, how are the zeros and poles
situated?
6. If R(z) is a rational function of order n, how large and how small
can the order of R'(z) be?
2. ELEMENTARY THEORY OF POWER SERIES
Polynomials and rational functions are very special analyt,ic functions.
The easiest way to achieve greater variety is to form limits. For
instance, the sum of a convergent series is such a limit. If the terms are
functions of a variable, so is the sum, and if the terms are analytic functions, chances are good that the sum will also be analytic.
Of all series with analytic terms the power series with complex
coefficients are the simplest. In this section we study only tKe most
elementary properties of power series. A strong motivation for taking
up this study when we are not yet equipped to prove the most general
properties (those that depend on integration) is that we need power series
to construct the exponential function (Sec. 3).
2.1. Sequences. The sequence (a,)? has the limit A if to every E > 0
there exists an no such that la, - AI < E for n 2 no. A sequence with a
finite limit is said to be convergent, and any sequence which does not cona, = a, the sequence may be said to
verge is divergent. If lim,,
diverge to infinity.
Only in rare cases can the convergence be proved by exhibiting the limit,
so it is extremely important to make use of a method that permits proof
of the existence of a limit even when it cannot be determined explicitly.
The test that serves this purpose bears the name of Cauchy. A sequence
will be called fundamental, or a Cauchy sequence, if it satisfies the following condition: given any E > 0 there exists an no such that \a, - a,\ < E
whenever n 2 no and m 2 no. The test reads:
A sequence i s convergent i f and only i f it i s a Cauchy sequence.
The necessity is immediate. If a, + A we can find no such that
(a, A ( < ~ / 2for n 2 no. For m , n 2_ no it follows by the triangle
inequality that (a, a,( 5 \a, A ( 4- (a, - A ( < E.
The sufficiency is closely connected with the definition of real numbers, and one way in which real numbers can be introduced is indeed to
postulate the sufficiency of Cauchy's condition. However, we wish to use
only the property that every bounded monotone sequence of real numbers has a limit.
-
-
-
34
COMPLEX ANALYSIS
The real and imaginary parts of a Cauchy sequence are again Cauchy
sequences, and if they converge, so does the original sequence. For this
reason we need to prove the sufficiency only for real sequences. We use
the opportunity to recall the notions of limes superior and limes inferior.
Given a real sequence {a,)? we shall set a, = max {al, . . , a,), that
is, a, is the greatest of the numbers all . . . , a,. The sequence {a,)? is
nondecreasing; hence it has a limit A I which is finite or equal to
W.
The number A I is known as the least upper bound or supremum (1.u.b. or
sup) of the numbers a,; indeed, it is the least number which is 1 all a,.
Construct in the same way the least upper bound Ak of the sequence
{ a , ) ~obtained from the original sequence by deleting all . . , ae-I.
I t is clear that {Ak)is a nonincreasing sequence, and we denote its limit
W , or - W .
I n any case we write
by A. I t may be finite,
.
+
.
+
A = lim sup a,.
m+ m
I t is easy to characterize the limes superior by its properties. If A is
E, and it follows
finite and E > 0 there exists an n o such that A,, < A
E for n 2 no.
In the opposite direction, if
that a, 5 A,, < A
a, 5 A - E for n h no, then A,, 5 A - E, which is impossible. I n
other words, there are arbitrarily large n for which a, > A - E. If
A = W there are arbitrarily large a,, and A = - W if and only if a,
tends to - W . I n all cases there cannot be more than one number A with
these properties.
The limes-inferior can be defined in the same manner with inequalities
reversed. I t is quite clear that the limes inferior and limes superior will
be equal if and only if the sequence converges to a finite limit or diverges
to
W or to - W .
The notations are frequently simplified to
and
lim. The reader should prove the following relations:
+
+
+
+
+ lim p,
5 lim (a,
lim a, + G 6, g G (a,
lim a,
+ P,)
+ P,)
5 lim an
5 E a,
+ p,
+ 5 p,.
Now we return to the sufficiency of Cauchy's condition.
+
From
Jan- anoJ< E we obtain Ia,I < lan,l E for n 2 no, and it follows that
A =
an and a = lim a, are both finite. If a + A choose
( A - a)
E =
3
and determine a corresponding no. By definition of a and A there exists
an a, < a E and an a, > A - E with m,n 2 n,. I t follows that
A - a = ( A - a,)
(a, - a,)
( a , - a) < 3 ~contrary
,
to thechoice
of E. Hence a = A, and the sequence converges.
+
+
+
COMPLEX FUNCTIONS
35
2.2. Series. A very simple applicat.ion of Cauchy's condition permits
us to deduce the convergence of one sequence from that of another. If it
is true that ib, - b,( 5 lam - a,! for all pairs of subscripts, the sequence
{b,] may be termed a contraction of the sequence {a,] (this is not a
standard term). Under this condition, if {a,] is a Cauchy sequence, so is
(b,] Hence convergence of { a , j implies convergence of { b,] .
An infinite series is a formal infinite sum
.
Associated with this series is the sequence of its partial sums
The series is said to converge if and only if the corresponding sequence is
convergent, and if this is the case the limit of the sequence is the sum of
the series.
Applied to a series Cauchy's convergence test yields the foilowing
condition: The series (15) converges if and only if to every E > 0 there
. . a,+,j < E for all n 2 no and
exists an no such that (a, a,+l
p >= 0. For p = 0 we find in particular that !a,\ < E. Hence the general term of a convergent series tends to zero. This condition is necessary,
but of course not sufficient.
If a finite number of the terms of the series (15) are omitted, the new
series converges or diverges together with (15). I n the case of convergence, let R, be the sum of the series which begins with the term a,+,.
Then the sum of the whole series is S = S ,
R,.
The series (15) can be compared wit,h the series
+
+ - +
+
+ +
.+
+
+
+ ...+
(16)
lall
lazl
la,l
.
formed by the absolute values of the terms. The sequence of partial
sums of (15) is a contraction of the sequence corresponding to (16), for
an+,\ 5 la,l
la,+ll
la,+,J. There(a,
a,+l
fore, convergence of (16) implies that the original series (15) is convergent.
A series with the property that the series formed by the absolute values
of the terms converges is said to be absolutely convergent.
+
+.- +
2.3. Unijorm Convergence. Consider a sequence of functions j,(x),
all defined on the same set E. If the sequence of values (f,(x)) converges for every X that belongs to E, then the limit f(x) is again a function
on E. By definition, if E > 0 and X belongs to E there exists an no such
that (j,(x) - ! ( X ) ( < E for n 11. no,but no is allowed to depend on X.
36
COMPLEX A N A L Y S I S
For instance, it is true that
+
for all X , but in order to have \ ( l l / n ) x - X \ = ( x l / n < E for n 2 no
it is necessary that no > IxJ/E.
Such an noexists for every fixed X , but
the requirement cannot be met simultaneously for all X .
We say in this situation that the sequence converges pointwise, but
not uniformly. In positive formulation: The sequence { fn(x) j converges
uniformly to f(x) on the set E i f to every E > 0 there exists an no such that
Ifn(x) - f i x ) ! < E for all n 2 no and all X in E.
The most important consequence of uniform convergence is the
following:
The limit function of a uniformly convergent sequence of continuous
functions i s itself continuous.
Suppose that the functions f n ( x ) are continuous and tend uniformly
to f ( x ) on the set E. For any E > 0 we are able to find an n such that
I f n ( x ) - f ( x ) < €13for all X in E. Let xo be a point in E. Because f n ( x )
is continuous at xo we can find 6 > 0 such that I f,(x) - f,(xo) 1 < e/3 for all
X in E with Ix - X,,[ < 6. Under the same condition on X it follows that
and we have proved that f(x) is continuous a t X O .
In the theory of analytic functions we shall find uniform convergence
much more important than pointwise convergence. However, in most
cases it will be found that the convergence is uniform only on a part of
the set on which the functions are originally defined.
Cauchy's necessary and sufficient condition has a counterpart for
uniform convergence. We assert :
The sequence f,,(x) ) converges uniformly on E if and only i f to every
F. > 0 there exists a n no such that If,(x) - f,(x) ( < E for all m , n 2 noand all
X in E.
The necessity is again trivial. For the sufficiency we remark that
the limit function f ( x ) exists by the ordinary form of Cauchy's test. I n
the inequality 1 f,(x) - fn(x)l < E we can keep n fixed and let m tend to
m. It follows that If ( x ) - f n ( x )( 5 E for n 2 no and all X in E. Hence
the convergence is uniform.
For practical use the following test is the most applicable: If a
sequence of functions { f n ( x ) ) is a contraction of a convergent sequence of
constants ( a , } , then the sequence { f n ( x )] is uniformly convergent. The
hypothesis means that Jf,(x) - fn(x)( 5 ( a , - an( on E, and the con-
COMPLEX FUNCTIONS
37
clusion follows immediately by Cauchy's condition.
In the case of series this criterion, in a somewhat weaker form, becomes
particularly simple. We say that a series with variable terms
has the series with positive terms
for a majorant if it is true that J f n ( x ) (5 M a n for some constant M and
for all sufficiently large n; conversely, the first series is a minorant of the
second. In these circumstances we have
Therefore, if the majorant converges, the minorant converges uniformly.
This condition is frequently referred to as the Weierstrass M test. I t has
the slight weakness that it applies only to series which are also absolutely
convergent. The general principle of contraction is more complicated,
but has a wider range of applicability.
EXERCISES
l.Prove that a convergent sequence is bounded.
2. If lim z, = A, prove that
n+
m
1
lim - (21
n+
,n
+ zz + . .
f 2,) = A-
3. Show that t,he sum of an absolutely convergent series does not
change if the terms are rearranged.
4. Discuss completely the convergence and uniform convergence of
the sequence { n z n )y.
5. Discuss the uniform convergence of the series
for real values of
X.
If U = u l + u : ! +
, V = v l + v ~ + . . - are convergent
series, prove that U V = u l v l + ( ~ 1 7 ~ 2 ~ 2 ~ 1 )( U I V ~ U ~ V Z U ~ U I ) . '
6.
"
+
+
+
+
+
provided that at least one of the series is absolutely convergent'. (It is
easy if both series are absolutely convergent. Try to arrange the proof so
economically that the absolute convergence of the second series is not
needed.)
38
COMPLEX A N A L Y S I S
2.4. Power Series. A power series is of the form
where the coefficients U , and the variable z are complex.
generally we may consider series
A little more
which are power series with respect to the center zo, but the difference is SO
slight that we need not do so in a formal manner.
As an almost trivial example we consider the geometric series
whose partial sums can be written in the form
Since zn -+ 0 for lzl < 1 and (znJ2 1 for lzl 2 1 we conclude that the
geometric series converges to 1/(1 - z ) for ( z ( < l, diverges for lzl 2 1.
I t turns out that the behavior of the geometric series is typical.
Indeed, we shall find that every power series converges inside a circle and
diverges outside the same circle, except that it may happen that the
series converges only for z = 0, or that it converges for all values of z.
More precisely, we shall prove the following theorem due to Abel:
Theorem 2. For every power series (17) there exists a number 8, 0 2
oo: called the radius of convergence, with the following properties:
(i) T h e series converges absolutely for every z with 1x1 < R. I f 0 S
p < R the convergence i s un<formfor J z (5 p.
(ii) If J z J> R the terms of the series are unbounded, and the series i s
consequently divergent.
(iii) I n lzl < R the sum of the series i s a n analytic function. T h e
derivative can be obtained by termwise diflerentiation, and the derived series
has the same radius of convergence.
R
5;
The circle J z J= R is called the circle of convergence; nothing is claimed
about the convergence on the circle. We shall show that the assertions in
the theorem are true if R is chosen according to the formula
1/R
=
-
n-
lim sup .\/lan(.
n+
COMPLEX FUNCTIONS
39
This is known as Hadamard's formula for the radius of convergence.
If ( z ( < R we can find p so that ( z ( < p < R. Then l / p > 1 / R , and
by the definition of limes superior there exists an no such that
< l/p,
(an( < l / p n for n 2 no. It follows that lanzn( < ((zl/p)"for large n, so t,hat
the power series (17) has a convergent geometric series as a majorant,,
and is consequently convergent. To prove the uniform convergence for
lzl 4 p < R we choose a p' with p < p' < R and find lanznl 5 (pip')" for
n 2 no. Since the majorant is convergent and has constant terms we
conclude by Weierstrass's M test that the power series is uniforqly
convergent.
If ( z ( > R we choose p so that R < p < 121. Since l / p < 1 / R there
are arbitrarily large n such that lan(l'n > l / p , (an(> l / p n . Thus
lanznl > ( ( ~ l / pfor
) ~infinitely many n, and the terms are unbounded.
m
2 nanzn-l has the same radius of convergence,
Proof: Set vi 1 + 6,. Then 6 , > 0, and by use
The derived series
1
because v n 3 1.
of the binomial theorem n = (1
6% < 2 / n , and hence 6 , 3 0.
For lzl < R we shall write
=
+ 6,)"
> 1 + 4 n ( n - 1)6%.
This gives
where
sn(z) = a0
+ alz + . + an-lzn-l,
m
Rn(z) =
2 akzk,
k=n
and also
m
f l(z) =
2 nanzn-l
=
1
We have to show that f'(z)
lim
n-+
S:
(2).
m
= fl(z).
Consider the identity
where we assume that z # zo and
rewritten as
(21, lzol
< p < R.
The last term can be
40
COMPLEX A N A L Y S I S
and we conclude that
The expression on the right is the remainder term in a convergent series.
Hence we can find nosuch that
1
n
-
n
- 20
5
o
1
3
for n 2 no.
There is also an nl such that Is;(zo) - fl(z0)l < ~ / 3for n 2 nl.
Choose a fixed n 2 no, n ~ . By the definition of derivative we can find
6 > 0 such that 0 < lz - zol < 8 implies
When all these inequalities are combined it follows by (19) that
when 0
< [ z - zol < 6.
We have proved that f'(zo) exists and equals
f1Go).
Since the reasoning can be repeated we have in reality proved much
more: A power series with positive radius of convergence has derivatives
of all orders, and they are given explicitly by
In particular, if we look at the last line we see that a,: = f("(O)/k!,and
the power series becomes
This is the familiar Taylor-Maclaurin development, but we have proved it
only under the assumption that f (z) has a power series development. We
do know that the development is uniquely determined, if it exists, but
the main part is still missing, namely that every analytic function has a
Taylor development.
COMPLEX FUNCTIONS
41
EXERCISES
l.Expand
(1 -. 2)-", m a positive integer, in powers of z.
2z+3.
2. Expand -in powers of z - 1. What is the radius of
z+l
convergence?
3. Find the radius of convergence of the following power series:
4. If Zanzn has radius of convergence R, what is the radius of convergence of ZanzZn?of Za:zn?
5. If f(z) = Zanzn,what is Zn3anzn?
6. If Zanznand Zbnznhave radii of convergence RI and R?, show that
the radius of collvergence of Zanbnznis a t least R1R2.
7. If liin,,
(an(/lan+ll= R, prove that Za,zn has radius of convergence R.
8. For what values of z is
$ (i%y
convergent?
9. Same question for
m
C
0
2.5. Abel's Limit Theorem. There is a second theorem of Abel's
which refers to the case where a power series converges at a point of the
circle of convergence. We lose no generality by assuming that R = 1
and that the convergence takes place a t z = 1.
m
Theorem 3.
If
2 a,
W
converges, then f(z) =
0
2 anzn tends to f(1)
as z
0
approaches 1 in such a way that 11 - zl/(l
- lzl) remains bounded.
Remark. Geometrically, the condition means that z stays in an angle
< 180" with vertex 1, symmetrically to the part (- a,, l ) of the real axis.
I t is customary to say that the approach takes place in a Stolz angle.
-
Proof.
We may assume that
2 an
0
=
0, for this can be attained by adding
COMPLEX ANALYSIS
42
a constant to ao. We write S, = a0
the identity (summation by parts)
+ a1 + . . . + an and make use of
But snzn-+ 0, so we obtain the representation
We are assuming that 11 - zl 5 K ( l - Ixl), say, and that S, -+ 0.
Choose m so large that lsnl < E for n 2 m. The remainder of the
series Zsnzn, from n = m on, is then dominated by the geometric series
(0
E
2 IzIn
m
=
E \ Z I ~ /-( ~(21) < €/(l - 121). It follows that
If(z)\ 6 11 - zl
m-l
/ 2
skzeI 4- KE.
0
The first term on the right can be made arbitrarily small by choosing z
sufficiently close to 1, and we conclude that f(z) -+ 0 when z --+ 1subject to
the stated restriction.
3. T H E E X P O N E N T I A L AND T R I G O N O M E T R I C F U N C T I O N S
'The person who approaches calculus exclusively from the point of view of
real numbers will not expect any relationship between the exponential
funct,ion eZand the trigonometric functions cos X and sin X. Indeed, these
functions seem to be derived from completely different sources and with
different purposes in mind. He will notice, no doubt, a similarity between
the Taylor developments of these functions, and if willing to use imaginary
arguments he will be able to derive Euler's formula eiz = cos X
i sin X
as a formal identity. But it took the genius of a Gauss to analyze its full
depth.
With the preparation given in the preceding section it will be easy to
define ez, cos x and sin z for complex x, and to derive the relations between
these functions. At the same time we can define the logarithm as the
inverse function of the exponential, and the logarithm leads in turn to the
correct definition of the argument of a complex number, and hence to the
nongeometric definition of angle.
+
3.1. The Exponential. We may begin by defining the exponential
function as the solution of the differential equation