Power series expansions

Many complex valued functions are represented by power series expansions.

We begin this section by recalling some of the results from earlier years about
innite series.
Convergence tests
The comparison test.
If |a
n
| b
n
for all n > N for some N, and if

b
n
converges, then

a
n
converges
absolutely.
The ratio test.
(a) If lim
n
|a
n+1
|/|a
n
| < 1. then

a
n
converges absolutely.
If the limit is > 1, then the series diverges. If the limit = 1, or if the limit does
not exist, the test is inconclusive.
(b)

a
n
(z z
0
)
n
converges absolutely if
lim
n
|a
n+1
||z z
0
|
n+1
|a
n
||z z
0
|
n
< 1
|z z
0
| < lim
n
|a
n
|
|a
n+1
|
providing the limit exists.
When it does,
R = lim
n
|a
n
|
|a
n+1
|
is called the radius of convergence of the power series.
Cauchys root test.
(a) if limsup|a
n
|
1/n
= q < 1, then

a
n
converges absolutely.
If q > 1, the series diverges. If q = 1, the test is inconclusive.
(b)

a
n
(z z
0
)
n
converges absolutely if
limsup|a
n
|
1/n
|z z
0
| < 1
|z z
0
| <
1
limsup|a
n
|
1/n
In this case, limsup|a
n
|
1/n
always exists, and the radius of convergence is
R =
1
limsup|a
n
|
1/n
However, Cauchys test is usually more docult to apply than the ratio test.
The Geometric Progression.
Consider
a + ar + ar
2
+ + ar
n1
=
n1

k=0
ar
k
1
2
If S = a + ar + ar
2
+ + ar
n1
then rS = ar + ar
2
+ + ar
n1
+ ar
n
S rS = (1 r)S = a ar
n
= a(1 r
n
)
S = a
1 r
n
1 r
If |r| < 1, then r
n
0 as n .

k=0
ar
k
=
a
1 r
If | z
0
| < |z z
0
|,
1
z
=
1
(z z
0
) ( z
0
)
=
1
z z
0
1
1
z
0
zz
0
=

k=0
1
z z
0
_
z
0
z z
0
_
k
=

k=0
( z
0
)
k
(z z
0
)
k+1
If | z
0
| > |z z
0
|,
1
z
=
1
z
=

k=0
(z z
0
)
k
( z
0
)
k+1
=

l=1
( z
0
)
l
(z z
0
)
l+1
Taylors Theorem
Suppose that f(z) is regular inside and on the circle C : |z z
0
| = R.
For any inside the circle,
| z
0
| < |z z
0
|
when z is on C, and Cauchys Integral Formula gives
f() =
1
2i
_
C
f(z)
z
dz
=
1
2i
_
C
f(z)

k=0
( z
0
)
k
(z z
0
)
k+1
dz
=

k=0
( z
0
)
k
_
1
2i
_
C
f(z)
(z z
0
)
k+1
dz
_
=

k=0
f
(k)
(z
0
)
k!
( z
0
)
k
3
Let M = max |f(z)| on C.
Using Cauchys inequalities,

f
(k)
(z
0
)

k!M
R
k
so that

k=0
f
(k)
(z
0
)
k!
( z
0
)
k

k=0

f
(k)
(z
0
)
k!

| z
0
|
k

k=0
M
_
| z
0
|
R
_
k
which converges absolutely for | z
0
| < R (at least).
For given z
0
, the radius of convergence of the Taylor series expansion is deter-
mined by the size of the largest circle within which f(z) remains regular.
That is, the radius of convergence is the distance from z
0
to the nearest singular
point of the function.
A function possessing convergent Taylor series expansions about every point in
a domain is said to be analytic.
This result shows that a regular function is analytic; i.e. in the complex plane,
the existence of the derivative at every point of a domain is sucient to guarantee
the existence of derivatives of all orders, and the convergence of the Taylos series
to the function in the domain.
e.g.
f(z) = log z
f

(z) =
1
z
; f

(z) =
1
z
2
; f

(z) =
2
z
3
f
(k)
(z) = (1)
k1
(k 1)!
z
k
log z = log z
0
+

k=1
(1)
k1
(k 1)!
k!z
k
0
(z z
0
)
k
= log z
0

k=1
1
k
_
1
z
z
0
_
k
This series converges provided

1
z
z
0

< 1
that is
|z z
0
| < |z
0
|
4
e.g.
f(z) = (1 + z)
p
f(0) = 1
f

(z) = p(1 + z)
p1
f

(0) = p
f(z) = p(p 1)(1 + z)
p2
f(0) = p(p 1)
f
(n)
(0) = p(p 1) . . . (p (n 1))
f(z) =

n=0
p(p 1) . . . (p n + 1)
n!
z
n
This is the Generalised Binomial Expansion.
If p is a positive integer, the series terminates, and we have the well-known
polynomial expansion.
Otherwise,
a
n
a
n+1
=
p(p 1) . . . (p n + 1)
n!
(n + 1)!
p(p 1) . . . (p n)
=
n + 1
p n

a
n
a
n+1

1 as n
Therefore the series converged for |z| < 1.
Analytic Functions
If
f(z) =

k=0
a
k
(z z
0
)
k
converges for |z z
0
| < R, then the series can be dierentiated term by term within
the circle of convergence, and
f

(z) =

k=1
ka
k
(z z
0
)
k1
and the derived series has the same radius of convergence as the original series.
This result shows that an analytic function is regular.
We have now shown that in the complex plane, regular, holomorphic and analytic
functions are the same. For this reason the labels are used interchangeably.
When
f(z) =

k=0
f
(k)
(z
0
)
k!
(z z
0
)
k
for |z z
0
| < R
0
, then
f

(z) =

k=1
f
(k)
(z
0
)
(k 1)!
(z z
0
)
k1
and
f
(n)
(z) =

k=n
f
(k)
(z
0
)
(k n)!
(z z
0
)
kn
also for |z z
0
| < R
0
.
5
If
f(z) =

k=0
a
k
(z z
0
)
k
then
f

(z) =

k=1
ka
k
(z z
0
)
k1
f

(z) =

k=2
k(k 1)(z z
0
)
k2
f
(n)
(z) =

k=n
k(k 1) . . . (k (n 1))(z z
0
)
kn
f(z
0
) = a
0
a
0
= f(z
0
)
f

(z
0
) = 1a
1
a
1
=
f

(z
0
)
1!
f

(z
0
) = 2.1a
2
a
2
=
f

(z
0
)
2!
f
(n)
(z
0
) = n.(n 1) . . . 1a
n
a
n
=
f
(n)
(z
0
)
n!
This shows that ANY convergent power series is the Taylor series.
Therefore we can use a variety of methods to obtain Taylor series expansions.
e.g. Starting with the expansion for e
z
, we have
e
z
=

k=0
1
k!
z
k
e
z
2
=

k=0
1
k!
(z
2
)
k
=

k=0
(1)
k
k!
z
2k
erf (z) =
2

_
z
0
e
s
2
ds
=
2

k=0
(1)
k
k!
z
2k+1
2k + 1
6
Similarly, starting from the Binomial expansion, we get
(1 + z)
p
= 1 + pz +
p(p 1)
2!
z
2
+ . . .
(1 z
2
)
1/2
= 1 +
_

1
2
_
_
z
2
_
+
1
2!
_

1
2
__

3
2
_
_
z
2
_
2
+ . . .
+
1
k!
_

1
2
__

3
2
_
. . .
_

2p 1
2
_
_
z
2
_
k
+ . . .
= 1 +
1
2
z
2
+
1
2!
1
2
3
2
z
4
+ . . .
+
1
k!
1
2
3
2
. . .
2k 1
2
z
2k
+ . . .
= 1 +
1
2
z
2
+
1.3
2
2
2!
z
4
+ . . .
+
1.3 . . . (2k 1)
2
k
k!
z
2k
+ . . .
=

k=0
(2k)!
2
2k
(k!)
2
z
2k
arcsinz =
_
z
0
ds

1 s
2
=

k=0
(2k)!
2
2k
(k!)
2
z
2k+1
2k + 1
Laurents Theorem
Suppose that f(z) is regular in the annular region bounded by C
1
: |zz
0
| = R
1
and C
2
: |z z
0
| = R
2
, where R
1
> R
2
.
If is in the annulus, then |z
0
| < |zz
0
| when z is on C
1
, and |z
0
| > |zz
0
|
when z is on C
2
.
Applying Cauchys Integral Formula to the annular region we have
f() =
1
2i
_
C
1
f(z)
z

1
2i
_
C
2
f(z)
z
=

k=0
(z z
0
)
k
_
1
2i
_
C
1
f(z)
(z z
0
)
k+1
dz
_
+

l=1
(z z
0
)
l
_
1
2i
_
C
2
f(z)
(z z
0
)
l+1
dz
_
=

a
n
(z z
0
)
n
This expansion is known as a Laurent expansion.
7
For given z
0
, the expansion is unique for the particular annular region under
consideration, but the same function may have dierent expansions in dierent
annuli about z
0
.
For example, consider the function
f(z) =
1
(z 1)(z 2)
=
1
z 2

1
z 1
.
For |z| < 1, we have the Taylor series expansions
f(z) =
1
1 z

1
2
1
1 z/2
=

n=0
z
n

1
2

n=0
_
z
2
_
n
=

n=0
_
1 2
(n+1)
_
z
n
For 1 < |z| < 2, we can retain the previous expansion for 1/(z 2), but we need
to replace the expansion of 1/(z 1) with one which is valid for |z| > 1.
f(z) =
1
z
1
1 1/z

1
2
1
1 z/2
=

n=0
z
(n+1)

n=0
2
(n+1)
z
n
Finally, for |z| > 2, we have
f(z) =
1
z
1
1 1/z
+
1
z
1
1 2/z
=

n=0
z
(n+1)
+

n=0
2
n
z
(n+1)
=

n=0
(2
n
1) z
(n+1)
The Laurent expansion represents the function f as the sum of two functions;
f = g + h ; where
g(z) =

k=0
_
1
2i
_
C
1
f(z)
(z z
0
)
k+1
dz
_
(z z
0
)
k
is regular for |z z
0
| < R
1
,
and
h(z) =

k=1
_
1
2i
_
C
2
f(z)(z z
0
)
k1
dz
_
1
(z z
0
)
k
is regular for |z z
0
| > R
2
.
8
Singular Points
Suppose that the function f has an isolated singular point (where it is not
regular) at z
0
, and consider the local Laurent expansion
f(z) =

k=1
a
k
(z z
0
)
k
+

k=0
a
k
(z z
0
)
k
for 0 < |z z
0
| < R.
We classify the singular point at z
0
according to the form of the principal part

k=1
a
k
(z z
0
)
k
of this expansion.
If this part of the expansion vanishes identically, we say that the singularity at
z
0
is removable.
For example
sinz
z
=

n=0
(1)
n
z
2n
(2n + 1)!
has a removable singularity at z = 0.
If f(z) has a removable singularity, then f(z) is bounded in a neighbourhood of
z
0
.
Conversely, If f(z) is bounded by M in a neighbourhood of z
0
, by choosing a
contour |z z
0
| = lying within this neighbourhood, we have
|a
k
| =
1
2

_
f(z)(z z
0
)
k1
dz

1
2
M
k1
2 = M
k
so that the singularity is removable.
It is usual in this cases to remove the singularity at z
0
by dening
f(z
0
) = a
0
.
For example,
f(z) =
_
e
z
1
z
z = 0
1 z = 0
is an entire function whose Taylor series expansion about z = 0 is
f(z) =

n=0
z
n
(n + 1)!
If a
m
= 0, but a
k
= 0 for all k > m, we say that f has a pole of order m at
z
0
.
If f(z) has a pole of order m at z
0
, then f(z) = g(z)/(z z
0
)
m
, where g(z) is
regular in a neighbourhood of z
0
, and g(z
0
) = a
m
= 0.
Therefore there is a neighbourhood of z
0
in which |g(z)| >
1
2
|a
m
|, and within
this neighbouthood
|f(z)| >
1
2
|a
m
|
|z z
0
|
m
which diverges to innity as z z
0
.
9
Finally, if the principal part has an innite number of non-zero terms in the
expansion, f(z) has an essential singularity at z
0
.
Consider any complex number A and any real > 0.
Suppose that |f(z) A| > in any neighbourhood of z
0
.
Then
1
|f(z) A|
<
1

in this neighbourhood, so that g(z) = 1/(f(z) A) has a removable singularity at

z
0
.
Therefore
f(z) = A +
1
g(z)
which has at worst a pole at z
0
if g(z
0
) = 0.
Therefore our supposition is not valid.
If f(z) has an essential singularity at z
0
, then f(z) comes arbitrarily close to
every complex number in every neighbourhood of z
0
.
In fact it can be shown that, with possibly two exceptions, f(z) takes every value
in C in every neighbourhood of z
0
For example, the function e
1/z
has an essential singularity at z = 0.
For any A = 0, we can solve
e
1/z
= A
1
z
= log(A) = log |A| + i arg(A)
= log |A| + i( + 2n) for some
z =
log |A| i( + 2n)
(log |A|)
2
+ ( + 2n)
2
and for any > 0 we can nd a solution (in fact innitely many of them) with
|z| < by taking n suciently large.
Analytic continuation
Suppose that f(z) is an analytic function, such that
f(z) =

n=0
a
n
(z z
0
)
n
for |z z
0
| < R
0
.
For any z
1
such that |z
1
z
0
| < R
0
, we can use the power series and its derivatives
to determine f(z
1
), f

(z
1
), etc., and hence we can (in theory at least) determine
the coecients in the expansion
f(z) =

k=0
f
(k)
(z
1
)
k!
(z z
1
)
k
which converges for |z z
1
| < R
1
.
If R
1
= R
0
|z
1
z
0
|, this new series gives no further information about the
function, but if R
1
> R
0
|z
1
z
0
|, then this expansion extends the region in which
we can evaluate f.
This extension is called an analytic continuation of f.
Zeros of analytic functions
10
If a
0
= a
1
= = a
n1
= 0, but a
n
= 0, then
f(z) =

k=0
a
k
(z z
0
)
k
has a zero of order n at z
0
.
f(z) = (z z
0
)
n
g(z), where
g(z) =

k=0
a
k+n
(z z
0
)
k
.
g(z
0
) = a
n
= 0. The function g is continuous.
Therefore given =
1
2
|a
n
|, there is > 0 such that
|g(z) a
n
| <
1
2
|a
n
| for |z z
0
| <
|g(z)| >
1
2
|a
n
|
Therefore there is a neighbourhood |z z
0
| r in which g(z) is not zero. This
shows that the zeros of f are isolated.
In this neighbourhood, g

(z)/g(z) is regular.
Suppose that g(z) = 0 for |z z
0
| r at least.
Consider
1
2i
_
|zz
0
|=r
f

(z)
f(z)
dz
=
1
2i
_
n(z z
0
)
n1
g(z) + (z z
0
)
n
g

(z)
(z z
0
)
n
g(z)
dz
=
1
2i
_ _
n
z z
0
+
g

(z)
g(z)
_
dz
= n + 0
which is the number of zeros of f(z) inside this contour.
Now consider
1
2i
_
C
f

(z)
f(z)
dz
for an arbitrary scroc C, where f is regular inside and on C, and such that no zero
of f lies on C.
As a consequence of Cauchys theorem, we can replace C by a series of circles
around the isolated zeros of f inside C.
Hence
1
2i
_
C
f

(z)
f(z)
dz = the number of zeros of f inside C
11
On the other hand, if we choose an arbitrary starting point a on C,
1
2i
_
C
f

(z)
f(z)
dz =
1
2i
log(f(z))|
a
a
=
1
2i
(log |f(a)| + i arg(f(a))
log |f(a)| i arg(f(a)))
While log |f(a)| takes the same value at the start and the nish of the contour,
arg(f(a)) increases by a factor 2 each time the curve f(z) circles the origin in a
counter-clockwise direction as z moves along C.
Hence
1
2i
_
C
f

(z)
f(z)
dz =
1
2
V ar arg f(z)|
C
which is the number of times f(z) circles the origin as z goes around C.
This means that to determine the number of zeros of an analytic function f(z)
inside a closed curve z = C(t); a t b, we plot the graph of f(C(t)) for a t b,
and count how many times it circles the origin.
Rouches Theorem
Suppose that f(z) and g(z) are regular inside and on the scroc C.
Then, if |g(z)| < |f(z)| on C, f(z) and f(z) + g(z) have the same number of
zeros inside C.
Proof. The number of zeros of f + g inside C is equal to
1
2
V ar arg(f + g)

C
.
However
f + g = f
_
1 +
g
f
_
arg(f + g) = arg(f) + arg
_
1 +
g
f
_
V ar arg(f + g)|
C
= V ar arg(f)|
C
+ V ar arg
_
1 +
g
f
_

C
and on C, |g/f| < 1, so that 1 +g/f remains inside the unit circle centered at 1 as
z traverses C.
Therefore,
V ar arg
_
1 +
g
f
_

C
= 0
and
V ar arg(f + g)|
C
= V ar arg(f)|
C
as required.
12
e.g. Consider
z
7
5z
3
+ 12
On C : |z| = 1, choose f(z) = 12 and g(z) = z
7
5z
3
.

z
7
5z
3

|z|
7
+ 5|z|
3
= 6 < 12
on the unit circle, so that Rouches Theorem holds, and z
7
5z
3
+12 has as many
zeros inside the unit circle as 12, namely none.
On C : |z| = 2, choose f(z) = z
7
, and g(z) = 12 5z
3
.
On C, |z
7
| = 2
7
= 128, and
|12 5z
3
| 12 + 5|z|
3
= 12 + 40 = 52 < 128
so that z
7
5z
3
+ 12 has all 7 of its zeros inside |z| = 2.
Stability
The stability of linear dynamical systems depends on having all the roots of a
certain polynomial either
(a) In the left half complex plane for continuous systems
(b) Inside the unit circle for discrete systems.
For the second case, we can use the preceding theory as is.
For the rst case, we need a minor modication.
Since we are dealing with a polynomial, there are only a nite number of zeros,
and therefore for suciently large R, all the roots lie inside the circle |z| = R.
All the roots of the polynomial will lie in the left half plane if they lie inside the
semicircle bounded by the imaginary axis and the circular arc z = Re
i
,

2

3
2
.
Given the polynomial
f(z) = z
n
+ a
n1
z
n1
+ . . . a
o
we can write this as
f(z) = z
n
_
1 + a
n1
z
1
+ . . . a
0
z
n
_
and for |z| suuciently large
f(z) = z
n
(1 + (z)) where || << 1
Therefore, on the semicircular arc,
f(z) z
n
= R
n
e
in
arg(f(z) n ;

2

3
2
V ar(arg(f(z)) n
where the approximations approach equality as R .
13
If all the zeros of f lie in the left half plane, then the total variation in arg(f(z)
around the contour is 2n.
Therefore, in this case the variation in arg(f(z) along the imaginary axis is also
n.
If, in addition, the coecients a
i
of the polynomial are real, so that f(x) is a
real polynomial, then by symmetry the variation of the argument along the positive
imaginary axis will be half the variation along the whole imaginary axis.
Therefore, for such a real polynomial, all the roots will have negative real part
provided the variation of the argument of f(iy) as y varies from 0 to is
1
2
n.
This variation can be determined by plotting the graph of
(Re(f(iy)), Im(f(iy))) ; y > 0 .
For example, if
f(z) = z
3
+ 6z
2
+ 11z + 6
f(iy) = iy
3
6y
2
+ i11y + 6
and all the roots of this polynomial will lie in the left half plane provided the graph
_
6 6y
2
, 11y y
3
_
; y > 0
goes three-quarters of the way around the origin.
When y = 0, the graph is at (6, 0).
When y = 1, the graph is at (0, 10).
When y =

11, the graph is at (60, 0).

Furthermore, these are the only places where the graph crosses the axes.
As y , the graph goes to in the third quadrant.
Therefore the variation in the argument, which must be an integral multiple of
1
2
, is
3
2
.
This shows that all the roots of this polynomial lie in the left half plane.
(They are in fact 1, 2 and 3.)
This result is known in the Engineering literature as Nyquists criterion.
Lagranges Expansion
Lemma. If the analytic function w(t) has a simple zero at t = inside the scroc
C and no other zeros inside C, and if F(z) is analytic inside and on C, then
1
2i
_
C
F(t)
w

(t)
w(t)
dt = F() .
Suppose that z is dened implicitly as a function of by the equation
z = a + (z) ,
where (z) is an analytic function of z inside and on a circle C: |z a| = , and
(a) = 0.
14
If M = max|| on C, then by Rouches Theorem the equation has a unique
solution inside C for || < /M.
The function w(t) = t a (t) has a simple zero at t = z(). Therefore, for
any function F, analytic inside and on C, and for || < /M,
F(z) =
1
2i
_
C
F(t)
1

(t)
t a (t)
dt
=
1
2i
_
C
F(t)
t a
(1

(t))
1
1 (t)/(t a)
dt
=
1
2i
_
C
F(t)
t a
(1

(t))
_

n=0

n
_
(t)
t a
_
n
_
dt
(since |/(t a)| < (/M)M/ = 1 on C)
=
1
2i
_
C
F(t)
t a
dt
+

n=1

n
1
2i
_
C
F(t)
_

n
(t a)
n+1


n1

(t a)
n
_
dt
= F(a)

n=1

n
1
2i
_
C
F(t)
n
_
d
dt
_

n
(t a)
n
__
dt
= F(a) +

n=1

n
n!
(n 1)!
2i
_
C
F

(t)
n
(t)
(t a)
n
dt
(integrating by parts)
= F(a) +

n=1

n
n!
_
d
n1
dt
n1
(F

(t)
n
(t))
_
t=a
Cauchys Integral Theorem
This expansion is known as Lagranges expansion.
In particular, if F(z) = z,
z = a +

n=1

n
n!
D
n1

n
(a)
We can use this expansion to determine inverse functions.
Suppose w = f(z), and f

(z
0
) = 0. Then locally we have
w w
0
= f(z) f(z
0
)
=
f(z) f(z
0
)
z z
0
(z z
0
)
= (z)(z z
0
) where (z
0
) = f

(z
0
) = 0
z z
0
= (w w
0
)(z) where (z) = 1/(z)
=

n=1
(w w
0
)
n
n!
D
n1

n
(t)

t=z
0
15
For example,
if w = z
1
z
; z
0
= 1
=
z
2
1
z
=
z + 1
z
(z 1)
z 1 = w
_
z
z + 1
_
=

n=1
w
n
n!
D
n1
_
t
n
(t + 1)
n
_

t=1
z = 1 +
1
2
w +
1
8
w
2
+ . . .