Chapter 11: laurents Transform Flashcards
Definition 11
isolated singularity
A function f which is analytic on the punctured disc
D’ = {z ∈ C : 0 < |z − α| < R} but not on {z ∈ C : |z − α| < R} is said to have an
isolated singularity at α
“isolated singularity” = “bad point”
EXAMPLE
1/sin z
has singularities
1/sin z
has singularities at z = nπ for n ∈ Z and is analytic on {z : 0 < |z| < π} for
example.
EXAMPLE
1/ (z-1)(z+2) has singularities
1/ (z-1)(z+2) has singularities at 1 and at −2, and is analytic on the punctured discs
{z : 0 < |z − 1| < 3} and {z : 0 < |z + 2| < 3}.
Theorem 11.2 (Laurent’s Theorem)
Suppose that f has an isolated singularity at α
(so f is analytic on some punctured disc D0 = {z : 0 < |z − α| < R}). Then f can be
represented on D0
by a Laurent series about z = α, i.e.,
f(z) = Σ from n =-∞ to ∞ of
a_n (z-α)^n
for z in D’.
The laurent coefficients are given by
a_n
= (1/2πi)∫_Cr f(w)/(w − α)^(n+1) . dw,
where Cr : w = α + reit (0 ≤ t ≤ 2π) for any 0 < r < R.
for two sets analytic a_n….
∫_Cr1 f(w)/(w − α)^(n+1) . dw
The an are independent of r: if 0 < r1 < r2 < R, we know that the function
f(w)/(w−α)^(n+1) is analytic on the D0 and is therefore analytic between Cr1 and Cr2
By Theorem
9.1, we see that
∫_Cr2 f(w)/(w − α)^(n+1) . dw,
and so is independent of r
DEF 11.3 residue and laurents
The Laurent coefficient a−1 is called the residue of f at α. We write
a−1 = Res {f; α}
f f has a singularity at α and Cr is a positively oriented circle centred on α of radius r,
then
∫_Cr f(z)dz = 2πi Res {f; α}.
EXAMPLE:
Find the residue of sinh z/z^4 at the origin.
Let f(z)= sinh z/z^4
Then f has a singularity at 0 and that it is analytic on C \ {0}
for z ≠ 0
f(z) = sinh z/z^4
= [z + (z^3 /6) + (z^5/120) + … ]/z^4
=
1/z^3 + ((1/6)/z) + (z/ 120) +…
and so Res {f : 0} =1/6
EXAMPLE:
Find the Laurent series of 1/ z(z−1) about z = 1 giving an expression for the general
term. Where is this expansion valid?
diagram D* and point
Let g(z) = 1/z(z-1)
g is analytic in C \ {0, 1}. It has isolated singularities at
the points 0,1. It is analytic on the punctured disc D∗ = {z ∈ C : 0 < |z − 1| < 1}
about 1. Hence g(z) has a Laurent series expansion, in powers of z − 1 valid in D∗
.
Let w=z-1
For 0 < |z − 1| < 1, i.e. 0 < |w| < 1,
g(z)= 1/ z(z−1)
= (1/(z-1)) - 1/z
= (1/w) - (1/(1+w))
= (1/w) - Σ from n =0 to ∞ of (-1)^n w^n
= (1/(z-1)) - Σ from n =0 to ∞ of (-1)^n (z-1)^n
is the Laurent series expansion for g(z) valid in D∗
if isolated singularity
If the function f has an isolated singularity at α, then f(z) has a Laurent Series
expansion
f(z) = Σ∞ n=−∞ of a_n(z − α)^n valid for z in some punctured disc D0
centred on α.
if analytic
If f is analytic at z0, then f(z) has a Taylor Series expansion f(z) = Σ∞ n=0 of a_n(z −z_0)^n
valid for z in some disc ∆ centred on z0.
A REMOVABLE SINGULARITY.
Definition 11.4
If a_n = 0 for all n < 0,
(so that b_n = 0 for all positive integers n,
i.e. all the negative powers of (z − α) in the Laurent expansion have coefficient 0) then α
is called a removable singularity.
REMOVABLE SINGULARITY FORM
f(z) = Σ∞ n=-∞ of a_n(z −α)^n
Suppose that f has an isolated singularity at α. Then f(z) has a Laurent series expansion
Σ from n=1 to ∞of b_n/(z −α)^n +
Σ from n=0 to ∞of a_n(z−α)^n
valid in some punctured disc about α, where bn = a−n for all positive integers n
If the function f has a removable singularity at α, then f(α) is undefined
If we define f(α) by (or change the value of f(α) so that) f(α) = a0,
then we get a function which is defined on a disc centred at α given by a Taylor series, i.e.
an analytic function. We have, then, removed the “removable singularity” by defining (or
redefining) f(α).
EXAMPLE
sinz/z
has a singularity at 0 as it is not defined there.
But the function
f(z) =
{sinz/z if z≠0
{ 1 if z=0
is analytic on C and
f(z) = 1 - (z^2/6) + (z^4/120) - (z^6/7!) +…
is its Taylor series about 0, valid for all z ∈ C.
B ISOLATED ESSENTIAL SINGULARITY.
Definition 11.5
If an ≠ 0 for infinitely many n < 0, (so that infinitely many of the
coefficients bn are non-zero
i.e. there are infinitely many negative powers of (z −α) with
non-zero coefficient in the Laurent expansion) then f has an isolated essential singularity
at α.
e^(1/z) and sin(1/z)
are both analytic on the punctured disc
D_0 = {z ∈ C : 0 < |z|},
so z = 0 is an isolated singularity of e^(1/z) and sin(1/z)
for z ≠ 0
e^(1/z) = 1 + (1/(1!z) + (1/2!z^2) + (1/3!z^3) + …
sin(1/z) = (1/z) - (1/3!z^3) + (1/5!z^5) -….
thus both e^(1/z) and sin(1/z) have isolated essential singularities at the origin
Note The even negative powers of z in the Laurent expansion of sin(1/z)
about the origin
all have coefficient zero. However, there are still an infinite number of negative powers
with non-zero coefficient
C. POLE def 11.6
If we can find k ∈ N such that a_{−k} ≠ 0 and a_n = 0 for n < −k, then
f is said to have a pole of order k at α
b_k ≠ 0 and b_n =0 for ALL n bigger than k and Laurent series only contains finite # of neg powers of (z− α) with non-zero coefficients
The Laurent series looks like
[a_{-k} / (z − α)^k]
\+
[a_{-{k-1}} / (z − α)^{k-1}] +
....+
a_-1 / (z − α) + a_0 + a_1(z − α) + .....
The Laurent series
of
sinhz /z^4
looks like:
(POLE)
for z≠ 0
sinhz /z^4 = (1/z^3) + (1/6)/z + z/120 +…
so sinhz /z^4 has a pole of order 3 @ the origin
. Note that 0 is an isolated singularity as
the function is analytic on the punctured disc D0 = {z ∈ C : 0 < |z|}
Note that sinh z has a zero to order 1 at the origin and z
4 has a zero to order
sinhz /z^4
therefore has a zero of order −3 in some sense
THM 11.7
f pole order k when
The function f has a pole of order k at α if and only if f(z) can be
expressed in the form
f(z) = g(z)/(z − α)^k
in some punctured disc D’ = {z ∈ C : 0 < |z − α| < R}, (R > 0) where, g is analytic and
non-zero in the disc D = D’ ∪ {α}
THM 11.8 zero order k and pole
If the function f has a zero of order k at α, then 1/f
has a pole of order k at α.
COROLLARY 11.9
sero order and pole quotient functs
If the function f has a zero of order m at α and the function g has a zero of order n at α, then
(i) f/ g
has a pole of order (n − m) at α if n > m;
(i) f/g
has a removable singularity at α if n ≤ m
SIMPLE POLE
DEF 11.10
If k = 1, we say that α is a simple pole.
Find all the singularities in the complex plane of
1/(1+z^2)
1/(1+z^2) is analytic in C \ {± i} and so has isolated singularities at ± i.
1+z^2 = (z-i)(z+i)
and so (1+z^2) has has zeros of order 1 at the points ± i.
thus 1/(1+z^2) has simple poles at ± i.
Find all the singularities in the complex plane of
sinz/z^2
sinz/z^2 is analytic on C \ {0} and so 0 is an isolated singularity. Now
sin z has a zero of order 1 at the origin and z^2 has a zero of order 2 at the origin.
Hence by Corollary 11.9
sinz/z^2
has a simple pole at the origin.
Find all the singularities in the complex plane of
1/(e^z -1)
e^z =1 if and only if z = 2nπi, where n is an integer.
let f(z) =(e^z -1) then
f(2nπi) = 0 f'(2nπi) = [e^z]_z=2nπi = e^(2nπi) =1 ≠ 0
Thus f has zeros of order 1 at the points 2nπi(n ∈ Z).
Hence 1/f
has isolated singularities at the points 2nπi(n ∈ Z) and these are simple
poles by Theorem 11.8
Determine the singularity of cotz/z
at the origin.
Let h(z) = cotz/z = cosz/zsinz = (f(z)/g(z)) where f(z) = cos z , g(z) = z sin z .
function h is analytic on the punctured disc {z ∈ C : 0 < |z| < π} and
so h has a isolated singularity at the origin
sin z has a zero of order 1 at the origin and, therefore, g has a zero of order 2 at the origin
By theorem 11.8, 1/g
has a pole of order 2 at the origin. Hence h =f/g
has a pole of order 2 i.e. a double pole at the origin since f is analytic in C and f(0) = cos 0 ≠ 0
Explain how Laurent expansions are used to classify isolated singularities.
Find all the singularities in the complex plane of funct and classify them
find the residue at each of the singularities
cos(1/(z-1))
Let g(z) = cos(1/(z-1))
g is analytic in C\{1}. For z ≠ 1,
g(z) = cos(1/(z-1))
= 1 - (1/2!)( 1/ (z-1))^2 + (1/4!)( 1/ (z-1))^4 - …
is the Laurent expansion of g(z) about the isolated singularity at
1. valid in C{1} and
Res{g;1} =0
Explain how Laurent expansions are used to classify isolated singularities.
Find all the singularities in the complex plane of funct and classify them
find the residue at each of the singularities
zcos(1/(z-1))
Let h(z) = zcos(1/(z-1))
h is analytic in C\{1}. For z ≠ 1,
h(z) = [(z-1) +1]cos(1/(z-1))
= [(z-1) +1] [1 - (1/2!)( 1/ (z-1))^2 + (1/4!)( 1/ (z-1))^4 - …]
=
1 + (z-1) - (1/2!)(1/(z-1))^2 - (1/2!)(1/(z-1)) + - (1/4!)(1/(z-1))^4 + - (1/4!)(1/(z-1))^3-…
is the Laurent expansion of h(z) about the isolated singularity at
1. valid in C{1} and
Res{h;1} = -1/2! = -0.5
THM 11.1
Quick ways of calculating residues at poles
- Suppose that f has a pole of order k at α. Then
Res {f ; α} =
[1/(k − 1)!] limz→α of [d^{k−1}/dz^{k−1}] [(z − α)^k f(z)] - If f =g/h
where g and h are analytic at α and g(α) ≠ 0, h(α) = 0, h’(α) ≠ 0 (so h has a simple zero at α), then f has a simple pole at α and
This shortcut will work when f(z) is given by a formula involving a fraction with an obvious factor (z − α)^n
in the denominator.
In other cases you will need to find a Laurent expansion and pick out the
coefficient a_{−1}
Find the singularities in the complex plane of the following and calculate the residues at
each of them:
sinz/(z^10)
Let q(z) = sinz/z^10
the q is analytic in C except for an isolated singularity. Now
sin z has a zero of order 1 at the origin and z
10 has a zero of order 10 at the origin. From
Corollary 11.9, we see that q has a pole of order 9 at the origin
(no shortcut and no 11.1, 1) )
laurent expansion for z≠0
sinz/(z^10) = (1/z^9) - 1/3!z^7) + (1/5!z^5) - (1/7!z^3)) + 1/(9!z) - z/11! + ….
Res(q;0} = 1/9!
Find the singularities in the complex plane of the following and calculate the residues at
each of them:
1/(1+z^2)
Let f(z) = 1/(1+z^2)
We know that f is analytic in C \ {± i} and has simple poles at ± i
let
g(z) = 1 , h(z) = 1 + z^2
h’(z) = 2z
g(i) =1
h(i)=0
h’(i) =2i
thus
Res{f;i}= g(i)/h’(i) = 1/2i
Res{f;-i}= g(-i)/h’(-i) = -1/2i
by Theorem 11.11 part (2).
Find the singularities in the complex plane of the following and calculate the residues at
each of them:
1/(e^z - 1)
Let k(z) = 1/(e^z - 1)
k is analytic in C except for simple poles at the
points 2nπi (n ∈ Z)
Theorem 11.11 part (2):
Res(k; 2nπi} = [1/(d/dz) (e^z − 1)]_z=2nπi = [1/e^z]_z=2nπi = 1.
Find the singularities in the complex plane of the following and calculate the residues at
each of them:
e^z/(z-1)^2
Let m(z) = e^z/(z-1)^2.
Then m is analytic on C except for a singularity at the point 1.
(z-1)^2 e^-z has a zero of order 2 at z = 1 and so its reciprocal m has a double pole
at 1.
thm 11.11 part 1 with k=2 and α = 1 gives,
Res{m;1} = (1/(2-1)!) lim(z→1) of (d/dz)[(z-1)^2 [e^z/(z-1)^2]]
= limz→1
of d(e^z)/dz
= e.
Find the singularities in the complex plane of the following and calculate the residues at
each of them:
1/(1+z^2)^9
Let p(z) = 1/(1+z^2)^9
function p is analytic on C except for isolated
singularities at ± i.
(1 + z^2)^9 = (z + i)^9(z − i)^9 has zeros of order 9 at ± i, we see that 1/(1+z^2)^9 has poles
of order 9 at ± i, i.e. m has poles of order 9 at ± i . Using Theorem 11.11 part (1) with
k = 9 gives,
Res{p;i}
= 16!/ ((8!)^2 2^17 i)
Res{p;-i}
= 16!/ ((8!)^2 2^17 i)
