Post on 20-May-2022
p3383.3theResidue.tk# y✓open
.Det Let felt I G- Isa3) have Laurent expansion abt .z=a
f-(z) =nI⇒Cn CZ -a) h ,valid Ze Brca) .
Thenresidue off at a tented Res Cf
,
a) : E,
.
-
,open
Note Let felt ( Elsa3) have Laurent expansion abt .z=a
f-(z) =nI⇒Cn CZ -am ,valid Ze Brca) .
Let 8pct ) ÷ at reit,o ETE 2T ,
OL re R and 8# c Bp .la)n k
9. see So E ch Cz -al MIE f on V*.
•
a- k=- n
so by uniform convergence (Cor I. . 2.6, pl7), get
n
Eckfqcz -alkdzT.nu?#c2SqfHdz-.k---n(Ex I. 2.7 p 17
11 Spr CZ -asrkdz I { Oz, ,
,if k t - I
,if k=
-
I
C-,C 21T i )
a
← This formula isso
pestf.al-I.fm#Z)dZJ..hndaPTfshfffo@nI.
I'
If,
also f- has a pole of order m at a Lso fcz) =I Cal Z -a)Y
-) .
k= - m , then
Res Cf,a) =
'IFA# ! IIIT Cz -ajmfcz)
II.mdnitgtiemrem ' Etienne# =¥T⇒n*h< producing the cm - D ! > & thentake Linna of the (m- D -trues diff . powerserie
# WTF E,
"
Emt im .§
Open y'Useful Fact Let p, QEIHCG) and Zoe G. (WTF Red Eq , Zo)),XIf q Lzo ) = 0 but pczo) ¥0 and
q'
Ho) to,
then at Zo, Iq has a simple pole Cdet a pole of order 1) and
Res ( Iq , Zo) = P¥ .
q' (Zo)
Why true ?LTGBG
.
L handwriting warning q #g )
(1.) By considering the power series ofQEHCG) at Zo,since qCZot=O>⇒to
,
qcz) = ( Z - Zo ) g CZ)for some g EH (G) Stg C. Zo) to . Gog has zero@zoof multiplicity 1),-7② So
pcz, PYgczL and I e Hla) cot Pho)1- =-
- F O .
qcz) CZ -Zo) G gczo)
so Iq has a simplepole at Zo . ( ie pole order m't )
wi m =/
③sores leg,
⇒ E- '¥m⇒o.I [⇐ - zDmTf¥)Iim Cz -EdtPCI by 12) 1 ,⇒Zo q Cz)
= IFZo Pg¥z)
P ,g ECCE) PL Zo) p ( Zo )-1- --
gczo) =p q'Ho)
( r ) ⇒ of'CE ) = g CZ) t ( Z
-
Zo) g'Cz)
⇒ q' ( Zo) = glzo ) to
p 34 Residue Thm ( Thm Ii . 3. I ) TF-
If §→ opens connected ! v (•I
↳ ¥.?.ge#arh:YisiiEF7oin+sP...e. !
te H (G1 Ep , ,. . .
, ph})path
'
L- -
8 be a closed piecewise" in G)Sp , ,. .
>Pnl ,Then
:S,fczgdz = ziti aft
,
[ Res Cf, Pa)) Indy(Pk) .
PI LTGBG .Find O CE CR sit . for each ke { I , nm , n } :
GU l note④ The Laurent Expansion off abt Pk
.
f- Cz) : =a
calm Cz - pwn is valid in Bpl Cpg.)② The balls { Bp
,Lpga) :3hh=
,
are disjoint .
(3) 8"C [ Be Cpk))
'
ien ,fBdPaD'
zeal iz - Peel > e V-E -1 . .,n3••
so for each k =L , in .
, n ,the singular part of the LE off abt ph, ie
Sch :=nI→dnh) Cz - put corn . uniformly on [ Beeped?Define g
: G) Sp . . .. . ,Pn3 → Q by :
g CZ) : I f Cz) - §,
Sch)(z) .
Note g E H ( Gl Ep , ... ,pn3) Lbk f- & Sch) 's are > .
Ate for each k , ge H ( Bpl Cpa) )
since the'balls 9 D
'
n(Ph )3y!
,
are disjoint (by 12dg.
Claim Fix k est, . . . ,n3 . Then g
EH ( B"
Cpa) ) extends tog e HLB Lpp) .4)( f z e Br' l Ph)
f ,⇒ -Z
s Cz)g Cz) : =
j -- I
= I aim Cz - pay - SHH - ¥h s" 'Cz)
n = - so
Iµ÷÷÷÷÷i:¥¥¥i÷::÷:*.s! what.is.
in thepurple box defines a bolam. (thus
continuous) function on Bk (pee) .So limp
,
g ( Z) exists,set igcpz) := lziyypu.gl Z) .
LAnd ph is a remote
. sing . of gon Br Cpa) and age H CBR (Pa)).
Thus g extends from G) Epi , Pn} tog e HCG) .
Cauchy 's Thin for A - like sets(Thin I.2.12 p2o)⇒ , Sy gtz) dz = O
⇒ Sy fczydz = £,Sp s" (⇒ da
.(4)
Fix k. It,chew ez - pan Sth) iz) on8*14
II.jcnh)Iz-panda ] S
,SH) Czsdz
if nt -1 : I z'
- pay = Iz Z'
{ d
'
z -Pat de-0[if n = I °
. Indo tph) 9¥ Eti Sr F-putZ .
eggso SysMlHdz = CY 12 Thi ) Indy ( Ph) .
byso Sp f- IZ) d 2-
Ep ÷ ,
Res ( f,Pa) 121T i) Ind p ( Pa) .
pgI (5)
5
Deaf Let f : 112-5112 be St flan, pig
is Riemann integrable f¥.
The Cauchy principle value integral of tf is :
(P.
V) f: fox) dx %± tiny, fr! fox) dx ;
provided the limit exists .
Note-
( D 18 x dx DNE but CPV ) Igo xdx = O .
⑦ f even function ⇒ fnk faux = 2 for tix) dx ,
Ex ( typical Qual Question) ..Find the principle value of I
•
XZS ⇒ dx .
-a
using the Residue Theorem . .
Consider z6= - I ⇒It= @ilttzth)
f- Cz) =ZI2-6+1
°
LEZM
f- Cz) EHLE l Epo , pi , . ..
, Ps3 ) where pies are the" 6th roots of
-
l"
,N
''
e ' starlikeph
= ei ( Ite t "¥) for k-0,1,-
,5
.
(ie pw smooth
.
-
To apply Res .Then . ..need closed path thatdoesn't passthru thepals
( and also want to somehow pick up [- R
,R ] CIR
.
R•
¥ , H5* Let Rs ,
Pa ...
. Po ✓(KR)#
-
R .- R
Pg" f " Ps
Pg =-
c'
HR : co,IT] -3K us 89 G) = Reit
BY :C- R, R] →a ur 82k Its = Reit-ter is the join of HR and ref =
''semicircle" Suk DX
-11
Ide Snr *Hdz = tzdz t {q ffzldzI
" Irma:3:* . Hofoper-38some"
nice " numberPV If tax
.
Apply Res .Them M HH = gf
2-left
{ p f# tz = ZITI k¥0 [ Res ( f, Pa) ] In¥%pI)iet
'r compute Res(f
, Pa) for k=0, 1,2 .
Res (Ey,
Pk) ="
1¥
f . gicpa,
"
= 6!h÷=tPI3 .Useful Fact Let p, QEIIIE) and Zoe G . (WTF RedEq , Zo))
¥¥¥÷÷÷÷÷÷÷÷:÷÷÷÷÷÷:÷.:¥¥÷÷.2-o=P k .
k> = t ( ⇐ it either)
- 3
=L e- it ⇐- i ;t£
-I2
= I C- i ) L -ok = if - 1)htt ( iz)
so,when R > I ,
fppftzdz = 2mi → C-Dhtt )) -4mi) (E) I - Itt - 1) =Iz .
Next try to show / Syr fcz, dz )R O.
'
p
¥?M often the ML Imax / length) Lemma is helpful.( Prop 2,5 ( 3 ) p 163 • Need te CC8*1 M 8 path .
I SatE)del a- [ mzaxey * Had ] KD
Recall €74 and if It) = Reit ,ostst
. I-
By the ML lemma
Isr.rE÷id4¥ecoiy max
' Hr,-
€,g*lI÷/ )n f :*, I. "' I'⇒ = :# Eso
.12-16 - IIIR> I% . use revere Ding
So
Spr ¥7 ,dz = Spa
,
II dz t Say IET dz
11 when Rsl fasho"
V
'
Iz o If 7¥,
dx.
so PRI: ¥. dx-detlpig.fr#adx=Iz.DEx .
Find Soo XI dx .
f-note fix) : III
,is even
Xbtl.
III. ↳ =
live.sc?IIi.idx-hi:..H.rII.dxwE.'zlEoI--It.
ERevisit .'
Find PV 15 7¥,dx •
&
• Had fcz) = IIe,
E H E e l s poop, ,. .
, pg 3) wipe = e
"
! :#,piecewise smooth curve
• To apply Res.Then . ..
need closed path that doesn't passthru thepals( and also want to somehow pick up [
- R,R ] a IR
.
Let R > 1R•
¥ ..
.45*13.
..
.
.
. Po ✓(KR)#
-.- p,
"
t .
pq =- i
R →0
Then it was not too hard to show fair f-Hdz→ O•
↳ this is often the case when f is a rational functionand 8Th is a Envoirole .
• But sometime,when trying to find a closed path that doesn't
pass thru the pts. . where f- is not different id be,a nice
semicircle circle will not work , Another commonly usedpath to try
'
is :'
ut r 't .÷÷*
) Bither:datedso
•
THEY'
Sqrfczsdz'
o.
-
p ,
"
I 12
. Tthis "
rectangle" is
Pg =- I
met a good choiceforf- CZ) =ZI .
Z6+1