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.
