Example: 10.9 Write down all possible Laurent expansions of the function ƒ(2) = (z−1)(z-2) around the point zo 3. = First of all we can use partial fractions to write f(2)=2²2-Å· The function has poles at z = 1,2 so the regions of interest are |z3| < 1, 1 < |z − 3| < 2 and |z − 3| > 2. This will give three different Laurent expansions. We can write 2 2 2-2 2 3+1 2Σ(-1)^(z − 3)"

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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How did they get that partial fraction. When I did it I got fractions. 

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Rangle 1.4: La fall, Wis the Laurent expansion
BT
is analytic. It is the posible to proses L
in this ring. Ikl, le ce partial fue deption, we
bod
Note that it came the c of the Laurent series be
Example 10. bet atbWitte Latent expanden
aroud. We wet to peal
of X
of 2-2 with post exposats
it is tic in the hel-2 <2
In order to see fillon Lorne noin in this asulas, w
festa partial fraction depodtis sa wake
-1-(7)
which cover ke-21. Tha
A AL(A)
which conges for< 1., that is, |-> 1. Putting the
which is wild ke 1-2
Zeroes and Poles
We can clasly the ladeal deketin of alytic actions
Sup that) wanadal duty. Thes
1
do il pr
Lai
00)
lepole af onder the anglaive power in the
pracipal part la 2 E21 scalad a szuple pola
(1) analgicky the principal part contala
70-l
Examples
Example 10.6: The best has a movable dar
Example 10.7: The reaction fiz) – 4 baxa cible pole at
(¹4) S
Examples
Example 16. The form)
layak
2-344148
abeme La
We can write
Examples 10.a Wise ees of were latest caperolase of
Examples
as moestid si
akal, by malige" is powers of 1/0
which is covererke (2-3 c
il tuo principal part of the
For the other fraction, we can
AARD
which is 3)
Examples
AAAAA
which is cover-11
Fox-ci war that
For 14-32 mm
a will ge the
------
Fly, -3>2wtar
---
-----
----
10 of 12
complex10_ce1b37448f28a25c7b4bd42caa064028
Example: 10.9 Write down all possible Laurent expansions of
the function f(z) :
=
3.
(z−1)(z–2) around the point zo =
First of all we can use partial fractions to write ƒ(z) :
=
2
2-2
The function has poles at z = 1,2 so the regions of interest are
|z − 3| < 1, 1 < |z − 3| < 2 and |z − 3| > 2. This will give three
different Laurent expansions.
We can write
ਜਣੇ
=
2
a
2
z−3+1
(-1)" (z − 3)"
2
1
2-2 2-1
Transcribed Image Text:Done Rangle 1.4: La fall, Wis the Laurent expansion BT is analytic. It is the posible to proses L in this ring. Ikl, le ce partial fue deption, we bod Note that it came the c of the Laurent series be Example 10. bet atbWitte Latent expanden aroud. We wet to peal of X of 2-2 with post exposats it is tic in the hel-2 <2 In order to see fillon Lorne noin in this asulas, w festa partial fraction depodtis sa wake -1-(7) which cover ke-21. Tha A AL(A) which conges for< 1., that is, |-> 1. Putting the which is wild ke 1-2 Zeroes and Poles We can clasly the ladeal deketin of alytic actions Sup that) wanadal duty. Thes 1 do il pr Lai 00) lepole af onder the anglaive power in the pracipal part la 2 E21 scalad a szuple pola (1) analgicky the principal part contala 70-l Examples Example 10.6: The best has a movable dar Example 10.7: The reaction fiz) – 4 baxa cible pole at (¹4) S Examples Example 16. The form) layak 2-344148 abeme La We can write Examples 10.a Wise ees of were latest caperolase of Examples as moestid si akal, by malige" is powers of 1/0 which is covererke (2-3 c il tuo principal part of the For the other fraction, we can AARD which is 3) Examples AAAAA which is cover-11 Fox-ci war that For 14-32 mm a will ge the ------ Fly, -3>2wtar --- ----- ---- 10 of 12 complex10_ce1b37448f28a25c7b4bd42caa064028 Example: 10.9 Write down all possible Laurent expansions of the function f(z) : = 3. (z−1)(z–2) around the point zo = First of all we can use partial fractions to write ƒ(z) : = 2 2-2 The function has poles at z = 1,2 so the regions of interest are |z − 3| < 1, 1 < |z − 3| < 2 and |z − 3| > 2. This will give three different Laurent expansions. We can write ਜਣੇ = 2 a 2 z−3+1 (-1)" (z − 3)" 2 1 2-2 2-1
3)
|
f(2)
(B)
1-2
18(2) = 2
2A + B = AZ-2)+B(z-1) 20:3
12-1/22) + Z-2 A2-2A+BZ-B- - 24-B=O→ -2A+H-A=o
21
A=L
(A+B=1-2B=#1-A - 1 - = =-=-=-=-=-=-
3
0
구
1
1-7-3)
_+ 21
13 21 32-2
A®
①
++칼(토) 글 치료께
기
3 2-1
ㅎ
ㅎㅇ
4
Transcribed Image Text:3) | f(2) (B) 1-2 18(2) = 2 2A + B = AZ-2)+B(z-1) 20:3 12-1/22) + Z-2 A2-2A+BZ-B- - 24-B=O→ -2A+H-A=o 21 A=L (A+B=1-2B=#1-A - 1 - = =-=-=-=-=-=- 3 0 구 1 1-7-3) _+ 21 13 21 32-2 A® ① ++칼(토) 글 치료께 기 3 2-1 ㅎ ㅎㅇ 4
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