Brewster H.D. Mathematical Physics
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172
Complex Representations
of
Functions
i
cosz dz = r
fez)
dz
=
2rrif(1)
=1=4
(=
-1)(z
-
5)
~=1=4
(z
-1)
Therefore, we have
i
cosz
dz
= rri
COS
(1)
=1=4
(=
-1)(=
-
5)
2
We have shown
that
j (zo) has an integral representation for
j{z)
analytic
in
Iz
-
zol
<
p.
In
fact, all derivatives
of
an analytic function have
an
x
i
cosz
Fig. Circular Contour used
in
Computing
=1=4
=2
_
6=
+ 5
d=
integral representation. This
is
given by
17!
cf
j(z)
j
(n)
(z)
=-
o 21ti
C(z-l)(z-5)
This can be proven following a derivation
similar
to
that for
the
Cauchy Integral Formula. One needs to recall the coefficients
of
the Taylor
series expansion for
j{z)
are given by
fn)(z
)
c =
O.
n n!
We also need the following lemma
Lemma:
ri
dz
{O,
':fc
(z -
zot+
1
= 21ti, n = °
This will be a homework problem. The integrals are similar to the n
= ° case above.
Geometric Series
In
this section we have made use
of
geometric series. A geometric
series
is
of
the form
00
I
ar
n
= a +
ar
+
ar2
+ ar2 +
...
+ arz +
....
n=O
.t
Complex
Representations of Functions 173
Here a is the first term and r
is
called the ratio. It is called the ratio because
the ratio
of
two
consecutive terms in the
sum
is
r.
The
sum
of
a geometric
series, when it converges, can easily be detennined.
We
consider the nth partial
sum.
Sn
= a +
ar
+
...
+
QI,J7-2
+
arl-1.
Now,
multiply
this
equation by
r.
1'S
I1
=
ar
+
a,.2
+
...
+
a,.n-1
+
ar
1
.
Subtracting these
two
equations, while noting the many cancellations,
we
have
(1
- 1')5
11
= a - QI,J7.
Thus, the nth partial
sums
can be
written
in the
compact
form
a(1-rl1)
S =
11
1-1'
Recalling that the sum,
if
it exists, is given by S = lim
l1
-t
oo
sl1'
Letting
n get large
in
the partial sum we need only evaluate lim
l1
-too
,.n.
From our
special limits
we
know that this limit is zero for
11'1
<
O.
Thus, we have the
sum
of
the geometric series.
CJ)
'"
ar
ll
--
~
L..
11'1
<
1.
11=0
1-,.
The
reader should verify that the geometric series diverges for all other
values
of
r.
Namely, consider what happens for the separate cases
11'1
>
1,
l'
= 1 and r = -
1.
Next, we present a few typical examples
of
geometric series.
",00
1
Example:
L..n=o
2n'
In
this
case
we
have
that
a = 1 and
l'
= 12. Therefore,
this
infinite
series converges and the sum is
1
S=-l
=
2.
1--
","00
4
Example: L..k=2
JT.
2
In
this
example
we
note
that
the
first
term
occurs
for
k = 2. So,
4 1
a =
'9'
Also, r
="3'
So,
S=
4
_9_=~
l-!
3
3
174
Complex
Representations of Functions
Example:
I:=l
(2:
-
s:
).
Finally, in this case we do not have a geometric series, but we do have
the difference
of
two geometric series.
Of
course, we need to be careful
whenever rearranging infinite series. In this case it is allowed. Thus, we
have
I
---
-I--I-
00(3
2)
00
3.
00
2
n=l
2
n
sn
-
n=l
2
n
n=!
sn .
Now
we can add both geometric series.
00
2 2
I(~-~)
"3
"5
1 S
n=!
2
n
Sn
=1_!-I_!=3-"2="2'
2 S
COMPLEX
SERI
ES
REPRESENTATIONS
Until this point
we
have only talked about series whose terms have
nonnegative powers
of
z -
zOo
It
is possible to have series representations
in
which
there are negative powers. In the last section we investigated
1 1
expansions
of
fez)
= 1 + z
about
z = 0
and
z
="2'
The
regions
of
convergence for each series. Let us reconsider each
of
these expansions,
but for values
of
z outside the region
of
convergence previously found.
1
Example: .f{z)
=-1
- for
Izl
> L
+z
As before, we make use
of
the geometric series. Since
Izi
>
1,
we
instead
rewrite
our
function as
1 2
.f{z)=
-=--
l+z
zl
l'
+-
z
We now have the function in a form
of
the sum
of
a geometric series
1
with first term a = 1 and ratio r
=--.
We note that
Izl
> 1 implies
that
z
Irl
<
1.
Thus, we have the geometric series
Complex
Representations of Functions
This can
be
re-indexed using j = n + 1 to obtain
<Xl
j(z) =
L(_I)j-1
z-<
j=l
175
Note that this series, which converges outside the unit circle,
Izl
>
1,
has negative powers
of
z.
Example:
j(z)
=l~Z
forlz-il>%·
In this case, we write
as
before
Instead
of
factoring out the % we factor out the ( z -
±)
term. Then,
we obtain
1
j(z)=-
l+z
Again, we identify a = 1 and r = - % ( z -
~
r I . This leads to the series
This converges for
I z -
~
I > % and can also be re-indexed to verify that
1
this series involves negative powers
of
z
-'2.
This leads to the following theorem.
Theore: Let
j(z)
be
analytic
in
an
annulus, Ri <
Iz
-
zol
< R
2
,
with C
a positively oriented simple closed curve around
zO
and inside the annulus.
Then,
<Xl
00
f(z)
=
La/z-zo)j
+
Lb/z-zor
j
,
j=O
j=i
176
Complex
Representations of Functions
iy
-1
1 x
-l
Fig. This figure shows an Annulus, R
J
<
Iz
-
zol
< R
2
,
with C a Positively Oriented
Simple closed Curve Around
Zo
and Inside the Annulus
with
and
=_l_A
fez)
dz
a
j
21ti
'-Jc
(z
-
zO)j+1
'
b.=_l_A
f(z).
dz
J 21ti'-JC(z-zO)-j+1 .
The
abO\
e series can
be
written
in
the more compact form
OCJ
j(z)
= I
cjCz-zo)j.
j=-oo
Such a series expansion
is
called a Laurent expansion.
Example: Expand
f{z) =
(1-
z)(2
+
z)
in
the annulus 1 <
Izl
< 2.
Using partial fractions, we can write this as
1
[1
1]
fCz)=-
-+-
.
3
l-z
2+z
1
We can expand the first fraction,
-1-
as an analytic function
in
the
-z
1
region
Izi
> 1 and the second fraction,
-2-'
as an analytic function
in
Izl
+z
< 2. This
is
done as follows. First, we write
Complex
Representations of Functions
Then we write
Therefore,
in
the common region, I <
Izl
< 2, we have that
I
[1
00
( z
)n
00
1 1
(l-Z)(2+Z)=3"2,~
-"2
-~Zn+l
I
oo
(-It
n I
oo
(-1)
-n
_
----z
+
---z
- n 3
n=O
6(2 )
n=O
.
SINGULARITIES
AND
THE
RESIDUE
THEOREM
177
In
the last section we found that we could integrate functions satisfying
some
analyticity
properties
along
contours
without
using
detailed
parametrizations around the contours.
We
can deform contours
if
the function
is
analytic
in
the region between the original and new contour.
In
this section
we will extend our tools for performing contour integrals.
The
integrand
in
th~
Cauchy
Integral
Formula
was
of
the
form
f(x)
I
g(z) =
----
Jf..z)
where
j(z)
is
well behaved at
zOo
The point z =
Zo
is
called a
z-zo
singularity
of
g(z), as g(z)
is
not defined there. As we saw from the proof
of'
the Cauchy Integral Formula, g(z) has a Laurent series expansion about z =
zO'
f(zo)
f'(
) 1
f"()(
)2
(
z)
=
--+
Zo
+-
Zo
z-zo
+ ...
g
z-zo
2
We now define singularities and then classify isolated singularities.
Definition: A singularity
ofj(z)
is
a point at whichj(z) fails to be analytic.
Typically
these
are
isolated singularities. In
order
to classify the
singularities
of
f (z),
we
look
at
the
principal
part
of
the
Laurent
series·I;=!
bj(z-zo)-j.
This
is
the part
of
the Laurent series containing
only negative powers
of
z -
z00
•
Iff(z)
is
bounded near
zo,
then
Zo
is
a removable singularity.
•
If
there are a finite number
of
terms
in
the principal part, then one
has a pole
of
order n
if
there
is
a term (z -
zotn
and no terms
of
the
form
(z - zo)-j for j >
n.
178 Complex Representations
of
Functions
•
If
there are an infinite number
of
terms in the principal part, then one
has an essential singularity.
sinz
Example: Removable.f{z)
=--.
z
At first it looks like there
is
a possible singularity at z =
O.
However,
sinz
we know from
the
first
semester
of
calculus
that
limz~o
--
=
1.
z
Furthermore, we can expand sin z about z = 0 and see that
sinz = .!.(z
-~+
... J =
1--=-=-+",
z z
6!
6!
Thus, there are only nonnegative powers in the series expansion. So,
this is an example
of
a removable singularity.
e
Z
Example: Poles .f{z) = (z
-It·
e
Z
For n = 1
we
have .f{z)
=--.
z-l
Z
For n = 1 we have .f{z)
=_e_.
This function has a singularity at z =
z-l
1 called a simple pole. The series expansion is found by expanding ez about
z = 1.
e
z-1 e e ( )
f{z) =
--e
=--+e+-
z-l
+ ...
z-l
z-I
2!
Note that the principal part
of
the Laurent series expansion about
z = 1 has only one term.
For
n = 2 we have .f{z) = (z _1)2 . The series expansion
is
found again
by expanding
ez about z =
1.
e z-1 e e e e
.f{
z)
=
(z-1)2
e = 2
+-+-+-(z-I)+
.. ·
(z-I)
z-l
2!
31
Note that the principal part
of
the
Laurent series has two terms
involving
(z - 1 )-2 and (z - 1 )-1. This
is
a pole
of
order
2.
1
Example: Essential f{z) =
-;.
e
Complex Representations
of
Functions 179
In this case
we
have the series expansion about z = 0 given
by
(
1
)n
1
00
-
001
j(z) =
e-;
=
I-z-=
I
_z-n
n=O
n!
n=O
n!
We
see
that
there
are
an
infinite
number
of
terms
in
the
principal
part
of
the
Laurent
series.
So,
this
function has
an
essential singularity at z =
O.
In the
above
examples
we
have
seen
poles
of
order
one
(a
simple
pole)
and
two
In
general,
we
can define poles
of
order
k.
Definition:
j(~)
has
~
pole
of
order k at
z~
if
and only
if
(z
-
zO)k
j(z)
has a removable smgularIty
at
zO'
but
(z -
zo)
-I
j(z) for k > 0 does not.
Let
<I>(z)
= (z -
zo)k
j(z) be analytic.
Then
it has a
Taylor
series
expansion
about
zOo
As
we
had
seen
in
equation
in the last section,
we
can
write
the integral
representation
of
the
(k
- 1 )st derivative
of
an analytic function as
(k-l)
~
=
(k
-1)!,{
$(z) dz
<I>
(':0)
2·
':fc (
)k·
m z -
Zo
Inserting the definition
of
<I>(z)
we
then have
<I>(k-l)
(zo)
=
(k
-
~)!,.r
f(z)dz
2m ':fc
Dividing
out
the factorial factor
and
evaluating the A(z) derivative,
we
have
that
lcf
lcf·
-.
f(z)dz
=
-.
f(z)dz
2nl
c 2nl c
1 d
k
-
l
k
=
(k
-I)!
dzk-I [(z - zo)
fez)
l=zo·
We note
that
from the integral representation
of
the coefficients for a
Laurent series, this gives
c_
1
'
or
hi.
This particular coefficient plays a role in helping
to
compute contour
integrals surrounding poles. It is called the residue
of
j(z)
at
z =
zOo
Thus,
for a pole
of
order k
we
define the residue
. 1 d
k
-
I
k
Res [f(z),
zo]
=
z~n:o
(k
-l)faz
k
-
I
[(z - zo)
fez)
]
Again, the residue is the coefficient
of
the (z -
zO)-1
term.
180
Complex
Representations of
Functions
C
Fig.
Contour
for Computing
~=I~l
s~z
Referring
to
the
last derivation, we
have
shown
that
if
fez) has
one
pole, zo'
of
order
k inside a simple closed
contour
C,
then
4
c
fCz)dz
=
27ti
Res[f(z),
zo]
q
dz
Example:
II
1
-.-
.
Z=
smz
We begin
by
looking
for
the
singularities
of
the
integrand,
which
is
when sin z =
O.
Thus, z = 0,
±7t,
±27t, ...
are
the
singularities. However, only z
= 0 lies inside
the
contour.
We
note
further
that
z = 0
is
a simple pole, since
lim(z-O)-.I-
=
1.
z~o
smz
Therefore,
the
residue
is
one
and
we
have
rf
dz =
27ti
'1jzl=l
sinz
.
In
general,
we
could
have
several
poles
of
different
orders.
For
example, we will
be
computing
rf
dz
'1jzl=l
z2
-1
.
The
integrand has singularities
at
z2
- 1 = 0,
or
z = ±
1,
both
poles are
inside the contour.
One could
do
a partial fraction decomposition
and
have
two integrals with one pole each. However, in cases in which we have many poles,
we can use the following theorem, known as the Residue Theorem.
Theorem:
Letf(z)
be
a
function
which has poles
zpJ
=
1,
... ,
~inside
a simple closed
contour
C
and
no
other
singularities
in
this region.
Then,
N
4c
fCz
)dz
=
27tiLRes[fCz);zj
J,
j=l
where
the
residues
are
computed
using
equation.
The
proof
of
this theorem
is
based
upon
the
contours.
One
constructs a
Complex Representations
of
Functions 181
new contour
C'by
encircling each pole.
Then
one
connects a
path
from C
to each circle.
In
the
figure two
paths
are shown only
to
indicate
the
direction followed
on
the
cut.
The new contour
is
then obtained by following C and crossing each cut
as
it
is
encountered.
Then one goes
around
a circle in the negative sense and returns along
the cut to proceed around
C.
The sum
of
the contributions to the contour
integration involve two integrals for each cut, which will cancel due to the
opposing directions. Thus,
we
are left with
gC,!(z)dz
=
4c
J
(z)dz-4c
1
J(z)dz
-
4C
2
J(Z)dZ-
4c
3
J(z)dz
= 0
Of
course, the sum
is
zero becausefiz)
is
analytic in the enclosed region,
since all singularities have be cut out. Solving
forgcJ(z)dz,
one has
that
this integral
is
the sum
of
the integrals around the separate poles, which can
be
evaluated with
single
residue
computations.
Thus,
the
result
is
that
gcJ(z)dz
is
2ni times the sum
of
the residues ofJ(z)
at
each pole.
A dz
Example:
':J]zl=2
z2
-1 .
We first note
that
there are two poles in this integral since
1
Z2
-1
= (z
-1)(z
+
1)
.
-2
2 x
-2i
Fig. A Depiction
of
how one cuts out Poles
to
prove
that
the Integral
Around
Cis
the sum
of
the Integrals
Around
Circles with the poles
at
the Centre
of
Each
We plot the contour
and
the two poles, each denoted by
an
"x". Since
both
poles are inside the contour,
we
need to compute the residues for each
one. They are
both
simple poles, so
we
have