complex varible
DESCRIPTION
TRANSCRIPT
direction. the on dependnot does value Its
exists ])()(
[lim)(
for ),(),()(
0
'
z
zfzzfzf
iyxzyxivyxuzf
z
yixzfyx
yixzfxy
yix
yxiyxyix
yix
ixyyxyyxxiyyxx
z
zfzzfzf
yixz
z
xyiyxzf
z
22)( 0 ,0 choose (2)
22)( 0 ,0 choose (1)
2)()(22
2))((2)()(
)()(lim)(
for
. of values allfor abledifferenti
is 2)( function thethat Show :Ex
'
'
22
2222
0
'
22
zzzz
zzz
z
zzzzf
ziyxzf
z
zz
22lim
]2)(
[lim])(
[lim)(
)()(
:method Another **
0
2
0
22
0
'
22
able.differenti nowhere is
)( so ,direction) (the on dependslimit The
1
2]
2[lim
)()(lim)(
slope of thriugh line a along 0 if
2222)()(
plane.complex the in anywhere abledifferenti
not is 2)( function thethat Show :Ex
0,0
'
zfm
im
im
yix
xiy
z
zfzzfzf
xmymzz
yix
xiy
yix
ixyxiixyy
z
zfzzf
ixyzf
yxz
direction. the oft independen is )(
that such everywhere analytic is )( ,1 Provided
)1(
1]
)1)(1(
1[lim
)]1
1
1
1(
1[lim]
)()([lim)(
.1at except
everywhere analytic is )1/(1)( function thethat Show :Ex
'
20
00
'
zf
zfz
zzzz
zzzzz
zfzzfzf
z
zzf
z
zz
20.2 Cauchy-Riemann relation
A function f(z)=u(x,y)+iv(x,y) is differentiable and analytic, there must be particular connection between u(x,y) and v(x,y)
relations Riemann-Cauchy - and
]),(),(),(),(
[lim
0 imaginary is suppose if (2)
]),(),(),(),(
[lim
0 real is suppose if (1)
]),(),(),(),(
[lim
),(),()(
),(),()(
])()(
[lim
0
0
0,
0
y
u
x
v
y
v
x
u
y
v
y
ui
yi
yxvyyxvi
yi
yxuyyxuL
xzx
vi
x
u
x
yxvyxxvi
x
yxuyxxuL
yz
yix
yxivyxuyyxxivyyxxuL
yyxxivyyxxuzzf
yixzyxivyxuzfz
zfzzfL
y
x
yx
z
function? analytic an ||||)(
is planecomplex the of domain which In :Ex
yixzf
|||]|[ (2)
quatrant second the 0y ,0 (b)
quatrant fouth the 0 ,0 (a) |]|[|| (1)
||),( |,|),(
xy
yxy
u
x
v
x
yxyy
xxy
v
x
u
yyxvxyxu
iy xiyx
zzf
zf
y
u
x
vi
y
v
x
u
z
y
y
f
z
x
x
f
z
f
izzyzzx
iyxzziyxz
not , of ncombinatio the
contains of fonction analytic an implies 0/satisfied. are
relations Riemann-Cauchy the then , analytic is )( If
)(2
)(2
1
2/)( and 2/)(
is of conjugatecomplex and
*
***
**
*
dimension. two in equation
sLaplace' of solutions are ),( and ),(
0 ),( functionfor result same the (2)
0 )()()()( (1)
satisfied are relations Riemann-Cauchy If
22
2
2
2
22
2
2
2
yxvyxu
y
v
x
vyxv
y
u
x
u
y
u
yx
v
yy
v
xx
u
x
orthogonal 0
ˆˆ),( and ˆˆ),(
are ly,respective curves, two the ingcorrespond vectors normal the
constant,),( andconctant ),( curves of families twoFor
x
u
y
u
y
u
x
u
y
v
y
u
x
v
x
uvu
jy
vi
x
vyxvj
y
ui
x
uyxu
yxvyxu
20.3 Power series in a complex variable
0at only converges 0 )!(lim ! (2)
allfor converges 0)!
1(lim
! (1)
edundetermin || (3)
divergent || (2)
convergent absolutely || (1) ||lim1
econvergenc of radius The
test".root Cauchy by" justisfied be can not,or convergent || Is
convergent absolutely is )( convergent is || if
)exp()(
/1
0
/1
0
/1
0
0
00
zRnzn
zRnn
z
Rz
Rz
RzaR
R
ra
zfra
inrazazf
n
nn
n
n
nn
n
nn
n
n
nn
n
nn
n
nn
n
nn
20.4 Some elementary functions
0 !exp Define
n
n
n
zz
terms. two above thefor of coeff. same the are There
!!
1
!!expexp
!!
1
!)!(
!
!
1
! is of coeff. the set
)...(!
1
!
)()exp(
)exp(expexpthat Show :Ex
21
210 000
21
21
22122
2122
11110
0
0
2121
2121
rs
rs
s rr
r
s
s
nrrs
nnn
rrnnr
nnnnnn
n
n
n
zz
zzrsr
z
s
zzz
rsrrn
n
nn
Czzsrn
zCzzCzzCzzCzCn
n
zzzz
zzzz
)sin)(cosexp( , if (3)
sincos...)!3
(.....!4!2
1
...!3!2
1)exp( if (2)
number real as same the exp)lnexp( if (1)
!
)(ln)lnexp(
0number real a ofcomonent complex the Define
342
32
0
yiyxeeeiyxzea
yiyy
yiyy
iyyiyeiyze,a
zezeea
n
azaza
a
iyxiyx
iy
z
n
nnz
)exp(
define we numbers,complex both are and 0 If
lnln
0 choosing by value principal its Take
. of function dmultivalue a is
)2(ln )]2(exp[
and real is for exp Write
expSet
zLntt
zt
-irz
n
zLnz
nirLnzwnirz
rirz
zw
z
])2(
exp[])2(
ln1
exp[
)]2(exp[ and )1
exp(
. of roots nthdistinct n exactly are therethat Show :Ex
11
1
n
kir
n
kir
nt
kirtLntn
t
t
nn
n
20.5 Multivalued functions and branch cuts
A logarithmic function, a complex power and a complex root are all multivalued. Is the properties of analytic function still applied?
)exp( and )( :Ex 2/1 irzzzf y
x
C
r
(A)
y
x
'C
r
(B)
(A)
circuit.
complete oneafter value original its to
return origin, the enclosenot dose
that Ccontour closed any traverse
z
(B)
2/1
2/1
2/12/1
)( function the ofpoint branch a is 0
)()(
)2/exp(
]2/)2(exp[)2/exp(
origin the enclose 2
zzfz
zfzf
ir
irir
Branch point: z remains unchanged while z traverse a closed contour C about some point. But a function f(z) changes after one complete circuit.
Branch cut: It is a line (or curve) in the complex plane that we must cross , so the function remains single-valued.
valued-single is )(
20restrict
)( :Ex 2/1
zf
zzf
y
x0
)()(2,2 then points, branch both (4)
)()(2, then ,not but (3)
)()(,2 then ,not but (2)
)()( , then point, branchneither (1)
encloses Ccontour If
)](exp[
)2/exp()2/exp()(
)exp( and )exp(set
:points branch expected ))((1)(
cuts. branch of tsarrangemen suitable
sketch hence and ,1)( of points branch the Find :Ex
2211
2211
2211
2211
2121
2121
2211
2
2
zfzf
zfzfiziz
zfzfiziz
zfzf
irr
iirrzf
iriziriz
izizizzzf
zzf
:follows ascut branch choose We.or either
containing loops around value changes )(
iziz
zf
y
i
i
x
(A) y
x
i
i
(B)
20.6 Singularities and zeros of complex function
.at order of pole a has )( the
,0)( and containing odneighborho some
in points allat analytic is )( integer, positive a is
)(
)()( :(pole) ysingularit Isolated
0
00
0
zznzf
zgzz
zgn
zz
zgzf
n
ysingularit essentiallimit the satisfies finite no if direction, any from (4)
as |)(|)( of pole a is if (3)
. thangreater order of pole a is then infinite, is if (2)
. than lessorder of pole a is then ,0 if (1)
number complex zero-non finite, a is and analytic is )(
)]()[(lim
is at order
of pole a has )(that for definition alternate An**
00
0
0
0
0
0
n
zzzfzfzz
nzza
nzza
azf
azfzz
zz n
zf
n
zz
1)(n pole simple a is ysingularit each
1}sinh
sinhcosh])2/1([{lim}
cosh
sinh])2/1([{lim
rule sHospital'l' Using
)2
1( )12(2
integer any is n )exp(])12(exp[exp
)exp(exp when ysingularit a has )(
)exp(exp
)exp(exp
cosh
sinh
tanh)( (2)
1 and 1at 1order of poles )1)(1(
2)(
1
1
1
1)( (1)
function the of iessingularit the Find :Ex
)2/1()2/1(
z
zzinz
z
zinz
inzniz
zniz
zzzf
zz
zz
z
z
zzf
zzzz
zzf
zzzf
inzinz
.approached is which from direction the oft independen and exists
)(limbut ed,undetermin )( of value the makes ySingularit
:essingularti Remove
0
0
z
zfzfzz
.0at ysingularit removable a has )(
so ,0 way the oft independen is 1)(lim
....!5!3
1........)!5!3
(1
)(
edundetermin 0/0)(lim :
0at ysingularit removable a is /sin)(that Show :Ex
0
5353
0
zzf
zzf
zzzzz
zzf
zfSol
zzzzf
z
z
zf
zf
/1 where ,0at )/1(
ofthat by given is infinityat )( ofbehavior The
zzf
nfzzf
z
fzzzf
zzf
bafzbzazf
zzfzzzf
bzazf
n
n
at ysingularit essential an has )(
)!()/1( exp)( (iii)
at 3order
of pole a has /1/1)/1( )1()( (ii)
at analytic is )( 0at analytic is
)/1( /1set )( (i)
exp)( (iii) and )1()( (ii)
)( (i) of infinityat behavior the Find :Ex
0
1
32
22
2
2
)(/1 of norder of pole a also is (iii)
zero. simple a called is 1,n if (ii)
n.order of zero a called is (i)
0)( and integer, positive a
is n if ,)()()( and 0)( If
0
0
0
0
00
zfzz
zz
zz
zg
zgzzzfzf n
20.10 Complex integral
dtdt
dxvidt
dt
dyuidt
dt
dyvdt
dt
dxu
vdxiudyivdyudx
idydxivudzzf
t
ttyytxx
tt
CCC C
C C
))(()(
is Bpoint
, is point A and )( ,)(
for ,parameter continuous real A2C
A 3C
y
x
1C
B
.at finishing and starting ,circle the
along ,/1)( of integralcomplex the Evaluate :Ex
RzR |z|
zzf
y
R
1C
x
t
. oft independen isresult calculated The
2sincos
cossin
by calculated also is integral The **
200
)sin)(sin
(coscos
cos)sin
()sin(cos1
sin ,
cos
,1
)( ,cos ,sin
20 ,sincos)(
2
0
2
0
2
0
2
0
2
0
2
0
2222
22
1
1
R
iidtdttiRtR
tiRtR
z
dz
iii
dttRR
titdtR
R
ti
tdtRR
tdttR
R
tdz
z
R
t
yx
yv
R
t
yx
xu
ivuyx
iyx
iyxzftR
dt
dytR
dt
dx
ttiRtRtz
C
C
3b3a3
2
C and C linesstraight two of up made Ccontour the (ii)
0 plane-half the
in || semicircle the of consisting Ccontour the (i)
along /1)( of integralcomplex the Evaluate :Ex
y
Rz
zzf
ca
xdx
xa
aii
titt
dttt
idttt
tdt
itt
i
dtiRRsiR
iRRdt
iRRtR
iRR
z
dz
sRsisRC
titRRtzC
izdzt
C
b
a
C
)(tan 2
)]2
(2
[2
0
|)]2/1
2/1(tan2[
2|)]221[ln(
2
1
221
1
221
12
1
1 term1st
)()(
10for )1( :
10for )1( : (ii)
/ 0
for but example, previous the in asjust is This (i)
122
10
110
2
1
0 2
1
0 2
1
0
1
0
1
0
3
3
3
2
y
R
bC3
1s
iR
aC3
R x
0t
path.different the oft independen is integral The
2)]
4(
4[0
|)2/1
2/1(tan|)]122[ln(
2
1
122
1
122
12
)1(
)]1()[1(
)1(
1 term 2nd
3
10
110
2
1
0 2
1
0 2
1
0 22
1
0
iz
dz
ii
siss
dsss
idsss
s
dsss
sisids
sis
i
C
examples. previous the in shown as C and C ,C path the
along )Re()( of integralcomplex the Evaluate :Ex
321
zzf
path.different the on depends integral The
)1(2
1)1(
2
1
)1()1)(1(
))(()()1(
:CCC (iii)2
)cossin(cos :C (ii)
)cossin(cos :C (i)
222
1
0
1
022
1
0
1
0
3b3a3
202
2
02
1
iRiRiR
dsisRdtitR
dsiRRsRdtiRRRt
Ri
dttiRtRtR
RidttiRtRtR
20.11 Cauchy theorem
C
dzzf
zfzf
0)(
Ccontour closed a on and withinpoint eachat
continuous is )( and function, analytic an is )( If '
apply. relations Riemann-Cauchy the and analytic is )(
0])(
[])()(
[
)()()(
and )(
)()(theorem divergence
ldemensiona-two by then C,contour closed a on
and within continuous are ),(
and ),(
If
zf
dxdyx
u
y
vidxdy
x
v
y
u
udyvdxivdyudxdzzfI
idydxdzivuzf
qdxpdydxdyy
q
x
p
y
yxq
x
yxp
RR
C C C
CR
.C and C along same the is )( of integral the then paths two
the by enclosed region the in and path each on function analytic
an is )( ifthat Show .C and C pathsdifferent two by
joined are planecomplex the in B and Apoints twos Suppose :Ex
21
21
zf
zf
A
1C
y
x
B
1 2
1 2 21
)()(
enclosingcontour closed a forms path
0)()()(
21
C C
C C CC
dzzfdzzf
RCC
dzzfdzzfdzzf
R
C
dzzfdzzfzf
C
C
)()( then contours two the between region the in analytic is )(
function the ifthat Show . with completely liesit that small lysufficient
being diagram, Argandthe in and contour closed twoConsider :Ex
y C
1C
2C
x
dzzfdzzfC
dzzfdzzfdzzfdzzf
dzzf
zf
C
CCC
)()( contour
ofthat as contour of direction the take If
)()()()(
0)(
analytic is )(
and , by bounded is area the
21
C
z fdzzfC
Rzzf
analytic. is )(0)(for , curve a by bounded
domain closed a in of function continuous a is )( if
:theorem sMorera'
20.12 Cauchy’s integral formula
Cdz
zz
zf
izfCz
Czf
000
)(
2
1)( then withinpoint a is and
contour closed a on and within analytic is )( If
C
)(2)(
)(
)exp( ),exp(for
)()(
0
02
0 0
2
00
0
00
zifdezfi
deie
ezfI
diidzizz
dzzz
zfdz
zz
zfI
i
ii
i
C
0z
)(
)(
2
1)(
:functioncomplex a of derivative the of form integral The
20
0' dz
zz
zf
izf
C
dzzz
zf
i
nzf
dzzz
zf
i
dzzzhzz
zf
i
dzzzhzzh
zf
i
h
zfhzfzf
C nn
C
Ch
Ch
h
10
0)(
20
000
000
00
00
'
)(
)(
2
!)( derivative nthFor
)(
)(
2
1
]))((
)(
2
1[lim
])11
()(
2
1[lim
)()(lim)(
.!
|)(|that show constant, some is
where circle, the on |)(| If .point the on centered
radius of circle a on and inside analytic is )(that Suppose :Ex
0)(
0
nn
R
MnzfM
MzfzzR
Czf
nnC nn
R
MnR
R
Mn
zz
dzzfnzf
!2
2
!|
)(
)(|
2
!|)(|
110
0)(
constant. a is )( then
allfor bounded and analytic is )( If :theorem sLiouville'
zfz
zf
zfzzfz
zzzf
zfzfRn
R
Mn!zf
nn
constant )( allfor 0)( plane.- the inpoint
any as take may we , allfor analytic is )( Since
0)(0|)(| let and 1set
|)(| :inequality sCauchy' Using
'
0
0'
0'
0)(
20.13 Taylor and Laurent series
n
n
nn
nn zz
n
zfzzazf
zzz
zf
)(!
)()()(
then C, insidepoint a is and ,point the on
centeredR radius of C circle a on and inside analytic is )( If
00
0)(
00
0
Taylor’s theorem:
!
)()(
!
)(2)(
2
1
)(
)()(
2
1)(
)(
2
1)(
)(11
in series geometric a as 1
expand
C on lies where )(
2
1)( so C, on and inside analytic is )(
0)(
00
0)(
00
100
00 0
0
0
0 0
0
00
0
n
zfzz
n
zifzz
i
dz
fzz
id
z
zz
z
f
izf
z
zz
zzz
zz
z
dz
f
izfzf
nn
n
nn
n
C nn
n
n
nC
n
n
C
202010
0
11
0
1
0
00
00
0
)()(.......)()(
)(
exp )( ,
.)()( series
Taylor a as expanded and at analytic is )()()( Then . on and
insidepoint other everyat analytic isbut at order of pole a has )( If
zzazzaazz
a
zz
a
zz
azf
eries Laurent sanded as a can bezfCinsidezallforThus
zzbzg
zzzfzzzgC
zzpzf
pp
pp
n
nn
p
021
0
10
10
10
0)(
on centered and circles two
between region a in analytic is )()(
)(
)(
2
1
)(
)(
2
1
)(
)(
2
1
!
)( and
zzCC
Rzzazf
dzzz
zf
idz
zz
zg
ia
dzzz
zg
in
zgbba
n
nn
npnn
n
n
npnn
0z
2C
1C
R
p
zfazzpzf
kaa
zzzf
zzm
zfmz-za
na
nazzzf
kpp
mm
n
n
ysingularit essential of valuelowest a find to impossible (ii)
)( of residue the called is ,at order of pole a has )(
0 allfor 0but 0 find to possible (i)
at analyticnot is )( If (2)
.at order of
zero a have to said is )( ,0 with )( is term
vanishing-nonfirst the ,0for 0 happen mayIt
.0for 0 all then ,at analytic is )( If )1(
10
0
0
0
0
n
nn zzazf )()( 0
pole. eachat )( of residue the find and 3,order of pole
a is 2 and 1order of pole a is 0that verify Hence .2 and 0
iessingularit theabout )2(
1)( of seriesLaurent the Find :Ex
3
zf
zzzz
zzzf
.8/1 is 2at )( of residue the 3,order of pole a is 2
32
2
16
1
)2(8
1
)2(4
1
)2(2
1..
3216
1
8
1
4
1
2
1
...])2
()2
()2
()2
(1[2
1
)2/1(2
1)(
)2/1(2)2()2( 2set 2point (2)
1order of pole a is 0 ...32
5
16
3
16
3
8
1
...])2
(!3
)5)(4)(3()
2(
!2
)4)(3()
2)(3(1[
8
1
)2/1(8
1)(
0point (1)
2323
43233
333
2
323
zzfz
z
zzz
zf
zzzz
zz
zz
zzz
zzzzf
z
001
1
10
0
01
101
1
101010
202010
0
1
0
at residue )]}()[()!1(
1{lim)(
limit the Take
)()!1()]()[(
...)(.......)()()(
...)()()(
......)(
)(
0
zzzfzzdz
d
mazR
zz
zzbamzfzzdz
d
zzazzaazfzz
zzazzaazz
a
zz
azf
mm
m
zz
n
nn
mm
m
mmm
m
mm
How to obtain the residue ?
)(
)(
)(
1lim)(
)(
)(lim)(
)(
)()(lim)(
0)( and at zero-non
and analytic is )( ,)(
)()( and at simple a has )( If (2)
)]()[(lim)(1 pole simple aFor (1)
0'
0'0
00
00
00
0
00
000
0
zh
zg
zhzg
zh
zzzg
zh
zgzzzR
zhz
zgzh
zgzfzzzf
zfzzzRm
zzzzzz
zz
.point theat )1(
exp)( function the of residue the
evaluate Hence .at )( of )( residue thefor expression
general a deriving ,about )( of seriesLaurent the gconsiderin
By .point theat order of pole a has )(that Suppose :Ex
22
00
0
0
izz
izzf
zzzfzR
zzf
zzmzf
e
ie
ie
i
iiR
iziz
iziz
i
iz
iz
dz
dzfiz
dz
d
iz
iziziziz
iz
z
izzf
2]
)2(
2
)2([
!1
1)(
exp)(
2exp
)(]
)(
exp[)]()[(
:at polefor
and at 2order of poles )()(
exp
)1(
exp)(
13
12
3222
2222
20.14 Residue theorem
0z
C
C
ninnin
nin
mnn
nC
mnn
ii
C
n
mnn
iadzzfI
iidn
ni
eidein
deiadzzzaI
deidzezz
dzzfdzzfI
zzazf
zzmzf
1
2
0
20
)1(1)1(2
01
)1(2
01
0
0
0
0
2)(
2 1for
0|)1(
1for
)(
set
)()(
)()(
at order of pole a has )(
Residue theorem:
C within poles itsat )( of residues the of sum the is
2)(
C within poles ofnumber finite afor except analytic, and
Ccontour closed a on and within continuous is )(
zfR
Ridzzf
zf
jj
jjC
C 'C
Ccontour open an along )( of integral The zfIC
0z
112
121100
0
101
0
2010
0
101
0
2 2contour closed afor
)()1
(lim)(lim
0)(lim
)()()(
except )( of ysingularit nothat enough small chosen is
)arg( and ||
gsurroundinneighbour some within analytic is )(
)()()(
at pole simple a has )( if
2
1
iaI
iadeie
adzzfI
dzz
dzzzadzzdzzfI
zzzf
zzzz
zz
zzazzf
zzzf
iiC
C
CC C
20.16 Integrals of sinusoidal functions
dzizdz
ziz
z
izdF
1
2
0
),1
(2
1sin ,)
1(
2
1cos
circleunit in expset )sin,(cos
0for cos2
2cos Evaluate :Ex
2
0 22
abd
abbaI
dz
ab
zba
zz
z
ab
i
ab
ba
zzz
dzz
ab
i
ababzzbzaz
idzz
zzabba
dzizzzd
abba
zzzzn nn
))((
)1(
2)1)((
)1(
2
)(
)1(21
)(21
2
))((21
cos2
2cos
)(2
12cos )(
2
1cos
2
4
22
4
2222
4
122
122
22
22
)(
2]
)([
22
)(
)(
)//()/(
1)/(]
)/)(/(
1)/[(lim)/(
1 ,/at pole (2)
//})/()/(
)]//(2)[1)(1(
)/)(/(
4{lim
]})/)(/(
1[
!1
1{lim)0(
2 ,0at pole (1)
)]()[()!1(
1{lim)( : Residue
circleunit the within / and 0at poles double ))((
1
2
222
2
22
4422
22
44
2
4
2
4
/
22
43
0
2
42
0
01
1
0
2
4
0
abb
a
abab
ba
ab
ba
ab
iiI
abab
ba
abbaba
ba
abzbazz
zbazbaR
mbaz
abbaabzbaz
abbazz
abzbaz
z
abzbazz
zz
dz
dR
mz
zfzzdz
d
mzR
bazzdz
ab
zba
zz
z
ab
iI
baz
z
z
mm
m
zz
C
20.17 Some infinite integrals
dxxf )(
. as zero to tends
along integral the ,) on || of maximum(2|)(|for
2)(
exist both )( and )( (3)
).|| as 0)(that is
condition sufficient (a as zero to tends on ||
of maximum the times , radius of semicircle a on (2)
axis. real the on is which of none poles, ofnumber finite a
forexcept ,0Im plane,-halfupper the in analytic is )( (1)
:properties following the has )(
0
0
R
ΓfRdzzf
Ridxxf
dxxfdxxf
zzzf
RΓf
RRΓ
zzf
zf
jj
y
R 0 R x
real is )(
Evaluate :Ex0 422
aax
dxI
77770 422
73
41
4442422
422
422422422
422422422
32
5
32
10
2
1
32
10)
32
5(2
)(
32
5)
2(
!3
)6)(5)(4(
)2(
1 is oft coefficien the
)2
1()2(
1
)2(
1
)(
1 0 ,set
plane-halfupper theat only
,at 4order of poles 0)(
)()(0
)(
as )()()(
aaI
aa
ii
ax
dx
a
i
a
i
a
a
i
aiaiazaiz
aiz
aizaz
ax
dx
az
dz
az
dz
Raz
dz
ax
dx
az
dz
CΓ
Γ
R
RC
R 0 R
ai
R 0 R - 0z
y
x
For poles on the real axis:
zero is integral the , as /1 thanfaster vanishes )( If
Re)(Re)(
Re Reset )(for (2)
)( at pole a has )(for (1)
)()()(
)()()()()(
contour closed afor
)()()(
0 as defined integral, the of value Principal
2
10
0
0
0
0
RRzf
difdzzf
didzzdzzf
aidzzfzzdzzf
dzzfdzzfdxxfP
dzzfdxxfdzzfdxxfdzzf
C
dxxfdxxfdxxfP
ii
iiθ
Γ
R
R
R
z
z
RC
R
z
z
R
R
R
Jordan’s lemma
contourar semicircul the is , as 0)(
then ,0 (3)
plane-halfupper the in || as 0|)(| of maximum the )2(
0Im in poles ofnumber
finite afor except plane-halfupper the in analytic is )( )1(
RdzzfeI
m
zzf
z
zf
imz
0 0 as
)1(2
0 ,on )(of maximum e
2
)(
)sinexp()exp(
2//sin1 ,2/0for
2/
0)/2(
2
0sin
0sin
IMRm
Me
m
MdeMRI
MRR |z||z |fM is th
dθeMR
deMR||dz|zf|eI
|θmR||imz|
mRmR
π/ θmR
Γ
π θmRimz
2/
)(f /2f
sinf
a R 0 R -
0 real, cos
of value principal the Find :Ex
madx
ax
mx
madxax
mxPmadx
ax
mxP
eaaidxax
eP
dzaz
eR
dzaz
edx
ax
edz
az
edx
ax
e
dzaz
eI
|z||az|
dzaz
eI
imaimx
imz
imzR
a
imximza
R
imx
C
imz
C
imz
cossin
and sincos
and 0
0 0 and As
0
as 0)( and plane,-halfupper
the in pole no 0 integral theConsider
11
1
20.18 Integral of multivalued functions
y
x
R
B A
D Copposite. and equalnot are
to joining CD and ABlines two along
values its d,multivalue is integrand The .0 and
let Weotigin. theat ispoint branch Single
, as such functions dMultivalue 2/1
Rzz
R
Lnzz
0for )(
:Ex0 2/13
axax
dxI
2/52/1
2
2
2/133
13
13
2/12/2/12/1
2/13
8
3
!2
1lim
])(
1)[(
)!13(
1lim)(
)( and ,at pole (2)
integralcontour the to oncontributi no make circles two the
and 0 as 0)( ,)()( integrand the (1)
a
iz
dz
d
zazaz
dz
daR
iaeaaaz
R|z|zfzazzf
az
az
i
2/50 2/13
2/50 2/13
2/5
0
, )22/1(2/132,0 2/13
20
2/5
8
3
)(
4
3
)()
11(
4
3
)()(
CD line along , ABline along
0 and 0 and
)8
3(2
axax
dx
axax
dx
e
aexaxe
dx
xax
dx
xezxez
dzdz
a
iidzdzdzdz
i
DC iiBA
ii
DCAB
dx sin
)( Evaluate :Ex22
x
xxI
2122222221 2
1
2
1sinIIdz
z
ze
idz
z
ze
idz
z
zzC
iz
C
iz
C
ii
iziziz
ixix
R
ix
C
iz
iz
ee
zii
dzz
ze
idz
z
ze
idz
z
ze
i
dxx
xe
idx
x
xe
i
dxx
xe
idz
z
ze
iI
ze
I
22)Res(2
2
1
2
1
2
1
2
1
2
1
2
1
2
1
2
1
.at pole one only and , term the to due
plane-halfupper the on choosed iscontour the ,for (1)
222222
2222
22221
1
21
1
1C
R -R -
1
2
iiiziz
izixix
R
ix
C
iz
iz
iiiix
iiz
iiz
ee
ii
dzz
ze
idz
z
ze
i
dzz
ze
idx
x
xe
idx
x
xe
i
dxx
xe
idz
z
ze
iI
ze
I
eeedxx
xe
iI
ezii
dzzz
ze
i
ezii
dzzz
ze
i
dzR
22
)()2()
2
1(
2
1
2
1
2
1
2
1
2
1
2
1
2
1
at pole one only , term
the by plane-halflower the choose we ,for (2)
2)(
42
1
4)Res(
2
1
))((2
1
4)Res()(
2
1
))((2
1
0 and 0 As
2222
222222
22222
2
221
2
1
2
2
1
1
2
2C
cos)(222
)(42
)(42
2
1
2
1sin)(
)(4
1
22
1
2)(
42
1
42))(
2
1(
))((2
1
42
)())(
2
1(
))((2
1
0 ,0 As
222222
22
222
2
1
iiiii
iiiiii
ixix
iiiix
iiiix
iiiz
iiiz
eeeee
eeeeee
dxx
xe
idx
x
xe
idx
x
xxI
eeedxx
xe
i
eeedxx
xe
iI
ee
ii
dzzz
ze
i
ee
ii
dzzz
ze
i
dzR