Chapter 20 Complex variables. 20.2 Cauchy-Riemann relation A function f(z)=u(x,y)+iv(x,y) is...

Chapter 20 Complex variables

direction. the on dependnot does value Its

exists ])()(

[lim)(

for ),(),()(

zfzzfzf

iyxzyxivyxuzf

yixzfyx

yixzfxy

yxiyxyix

ixyyxyyxxiyyxx

zfzzfzf

xyiyxzf

22)( 0 ,0 choose (2)

22)( 0 ,0 choose (1)

2)()(22

2))((2)()(

)()(lim)(

. of values allfor abledifferenti

is 2)( function thethat Show :Ex

ziyxzf

[lim])(

[lim)(

:method Another **

able.differenti nowhere is

)( so ,direction) (the on dependslimit The

)()(lim)(

slope of thriugh line a along 0 if

2222)()(

plane.complex the in anywhere abledifferenti

not is 2)( function thethat Show :Ex

zfzzfzf

xmymzz

ixyxiixyy

direction. the oft independen is )(

that such everywhere analytic is )( ,1 Provided

)1)(1(

1[lim]

)()([lim)(

.1at except

everywhere analytic is )1/(1)( function thethat Show :Ex

zfzzfzf

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

]),(),(),(),(

0 imaginary is suppose if (2)

]),(),(),(),(

0 real is suppose if (1)

]),(),(),(),(

),(),()(

yxvyyxvi

yxuyyxuL

yxvyxxvi

yxuyxxuL

yxivyxuyyxxivyyxxuL

yyxxivyyxxuzzf

yixzyxivyxuzfz

zfzzfL

function? analytic an ||||)(

is planecomplex the of domain which In :Ex

|||]|[ (2)

quatrant second the 0y ,0 (b)

quatrant fouth the 0 ,0 (a) |]|[|| (1)

||),( |,|),(

yyxvxyxu

iy xiyx

izzyzzx

iyxzziyxz

not , of ncombinatio the

contains of fonction analytic an implies 0/satisfied. are

relations Riemann-Cauchy the then , analytic is )( If

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

yxvyxu

orthogonal 0

ˆˆ),( and ˆˆ),(

are ly,respective curves, two the ingcorrespond vectors normal the

constant,),( andconctant ),( curves of families twoFor

yxvyxu

20.3 Power series in a complex variable

0at only converges 0 )!(lim ! (2)

allfor converges 0)!

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()(

inrazazf

20.4 Some elementary functions

0 !exp Define

terms. two above thefor of coeff. same the are There

!!expexp

! is of coeff. the set

)...(!

)()exp(

)exp(expexpthat Show :Ex

210 000

nnnnnn

Czzsrn

zCzzCzzCzzCzCn

)sin)(cosexp( , if (3)

sincos...)!3

(.....!4!2

...!3!2

1)exp( if (2)

number real as same the exp)lnexp( if (1)

)(ln)lnexp(

0number real a ofcomonent complex the Define

yiyxeeeiyxzea

iyyiyeiyze,a

zezeea

iyxiyx

define we numbers,complex both are and 0 If

0 choosing by value principal its Take

. of function dmultivalue a is

)2(ln )]2(exp[

and real is for exp Write

expSet

nirLnzwnirz

exp[])2(

)]2(exp[ and )1

. of roots nthdistinct n exactly are therethat Show :Ex

kirtLntn

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

circuit.

complete oneafter value original its to

return origin, the enclosenot dose

that Ccontour closed any traverse

2/12/1

)( function the ofpoint branch a is 0

)2/exp(

]2/)2(exp[)2/exp(

origin the enclose 2

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

)()(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

zfzfiziz

iirrzf

iriziriz

izizizzzf

:follows ascut branch choose We.or either

containing loops around value changes )(

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

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**

zzzfzfzz

1)(n pole simple a is ysingularit each

1}sinh

sinhcosh])2/1([{lim}

sinh])2/1([{lim

rule sHospital'l' Using

1( )12(2

integer any is n )exp(])12(exp[exp

)exp(exp when ysingularit a has )(

)exp(exp

tanh)( (2)

1 and 1at 1order of poles )1)(1(

1)( (1)

function the of iessingularit the Find :Ex

)2/1()2/1(

inzniz

inzinz

.approached is which from direction the oft independen and exists

)(limbut ed,undetermin )( of value the makes ySingularit

:essingularti Remove

zfzfzz

.0at ysingularit removable a has )(

so ,0 way the oft independen is 1)(lim

....!5!3

1........)!5!3

edundetermin 0/0)(lim :

0at ysingularit removable a is /sin)(that Show :Ex

/1 where ,0at )/1(

ofthat by given is infinityat )( ofbehavior The

bafzbzazf

zzfzzzf

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

)(/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

zgzzzfzf n

20.10 Complex integral

dxvidt

dyuidt

vdxiudyivdyudx

idydxivudzzf

ttyytxx

))(()(

is Bpoint

, is point A and )( ,)(

for ,parameter continuous real A2C

.at finishing and starting ,circle the

along ,/1)( of integralcomplex the Evaluate :Ex

RzR |z|

. oft independen isresult calculated The

2sincos

cossin

by calculated also is integral The **

)sin)(sin

(coscos

cos)sin

()sin(cos1

)( ,cos ,sin

20 ,sincos)(

iidtdttiRtR

titdtR

iyxzftR

ttiRtRtz

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

dtiRRsiR

sRsisRC

titRRtzC

)(tan 2

|)]2/1

2/1(tan2[

2|)]221[ln(

1 term1st

10for )1( :

10for )1( : (ii)

for but example, previous the in asjust is This (i)

path.different the oft independen is integral The

2/1(tan|)]122[ln(

)]1()[1(

1 term 2nd

sisids

examples. previous the in shown as C and C ,C path the

along )Re()( of integralcomplex the Evaluate :Ex

path.different the on depends integral The

)1()1)(1(

))(()()1(

:CCC (iii)2

)cossin(cos :C (ii)

)cossin(cos :C (i)

iRiRiR

dsisRdtitR

dsiRRsRdtiRRRt

dttiRtRtR

RidttiRtRtR

20.11 Cauchy theorem

Ccontour closed a on and withinpoint eachat

continuous is )( and function, analytic an is )( If '

apply. relations Riemann-Cauchy the and analytic is )(

[])()(

)()()(

and )(

)()(theorem divergence

ldemensiona-two by then C,contour closed a on

and within continuous are ),(

and ),(

vidxdy

udyvdxivdyudxdzzfI

idydxdzivuzf

qdxpdydxdyy

.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

1 2 21

enclosingcontour closed a forms path

0)()()(

C C CC

dzzfdzzf

dzzfdzzfdzzf

dzzfdzzfzf

)()( 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

dzzfdzzfC

dzzfdzzfdzzfdzzf

)()( contour

ofthat as contour of direction the take If

)()()()(

analytic is )(

and , by bounded is area the

z fdzzfC

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

1)( then withinpoint a is and

contour closed a on and within analytic is )( If

)exp( ),exp(for

zifdezfi

diidzizz

:functioncomplex a of derivative the of form integral The

dzzzhzz

dzzzhzzh

zfhzfzf

!)( derivative nthFor

)()(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

MzfzzR

nnC nn

dzzfnzf

constant. a is )( then

allfor bounded and analytic is )( If :theorem sLiouville'

zfzzfz

zfzfRn

constant )( allfor 0)( plane.- the inpoint

any as take may we , allfor analytic is )( Since

0)(0|)(| let and 1set

|)(| :inequality sCauchy' Using

20.13 Taylor and Laurent series

zfzzazf

)()()(

then C, insidepoint a is and ,point the on

centeredR radius of C circle a on and inside analytic is )( If

Taylor’s theorem:

in series geometric a as 1

expand

C on lies where )(

1)( so C, on and inside analytic is )(

202010

)()(.......)()(

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

eries Laurent sanded as a can bezfCinsidezallforThus

zzzfzzzgC

on centered and circles two

between region a in analytic is )()(

)( and

Rzzazf

zfazzpzf

zfmz-za

nazzzf

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(

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

.8/1 is 2at )( of residue the 3,order of pole a is 2

...])2

)2/1(2

)2/1(2)2()2( 2set 2point (2)

1order of pole a is 0 ...32

...])2

)5)(4)(3()

)4)(3()

2)(3(1[

)2/1(8

0point (1)

101010

202010

at residue )]}()[()!1(

1{lim)(

limit the Take

)()!1()]()[(

...)(.......)()()(

...)()()(

......)(

zzzfzzdz

zzbamzfzzdz

zzazzaazfzz

zzazzaazz

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)

zgzzzR

zgzfzzzf

zfzzzRm

zzzzzz

.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

zzzfzR

iziziziz

exp[)]()[(

:at polefor

and at 2order of poles )()(

20.14 Residue theorem

ninnin

iadzzfI

eidein

deiadzzzaI

deidzezz

dzzfdzzfI

)1(1)1(2

2 1for

at order of pole a has )(

Residue theorem:

C within poles itsat )( of residues the of sum the is

C within poles ofnumber finite afor except analytic, and

Ccontour closed a on and within continuous is )(

Ridzzf

Ccontour open an along )( of integral The zfIC

121100

2 2contour closed afor

(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

iadeie

adzzfI

dzzzadzzdzzfI

zzazzf

20.16 Integrals of sinusoidal functions

dzizdz

1sin ,)

circleunit in expset )sin,(cos

0for cos2

2cos Evaluate :Ex

ababzzbzaz

zzabba

dzizzzd

zzzzn nn

2)1)((

))((21

12cos )(

)//()/(

)/)(/(

1)/[(lim)/(

1 ,/at pole (2)

//})/()/(

)]//(2)[1)(1(

)/)(/(

]})/)(/(

1{lim)0(

2 ,0at pole (1)

)]()[()!1(

1{lim)( : Residue

circleunit the within / and 0at poles double ))((

abbaba

abzbazz

zbazbaR

abbaabzbaz

abbazz

abzbaz

abzbazz

zfzzdz

bazzdz

20.17 Some infinite integrals

dxxf )(

. as zero to tends

along integral the ,) on || of maximum(2|)(|for

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 )(

ΓfRdzzf

Ridxxf

dxxfdxxf

R 0 R x

real is )(

Evaluate :Ex0 422

77770 422

4442422

422422422

)6)(5)(4(

1 is oft coefficien the

1 0 ,set

plane-halfupper theat only

,at 4order of poles 0)(

as )()()(

aiaiazaiz

R 0 R - 0z

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

difdzzf

didzzdzzf

aidzzfzzdzzf

dzzfdzzfdxxfP

dzzfdxxfdzzfdxxfdzzf

dxxfdxxfdxxfP

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

0 0 as

0 ,on )(of maximum e

)sinexp()exp(

2//sin1 ,2/0for

MdeMRI

MRR |z||z |fM is th

dθeMR

deMR||dz|zf|eI

|θmR||imz|

π/ θmR

π θmRimz

)(f /2f

a R 0 R -

0 real, cos

of value principal the Find :Ex

madxax

mxPmadx

eaaidxax

|z||az|

imaimx

imximza

cossin

and sincos

0 0 and As

as 0)( and plane,-halfupper

the in pole no 0 integral theConsider

20.18 Integral of multivalued functions

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

0for )(

:Ex0 2/13

2/52/1

2/12/2/12/1

1lim)(

)( and ,at pole (2)

integralcontour the to oncontributi no make circles two the

and 0 as 0)( ,)()( integrand the (1)

iaeaaaz

R|z|zfzazzf

2/50 2/13

, )22/1(2/132,0 2/13

CD line along , ABline along

0 and 0 and

aexaxe

xezxez

iidzdzdzdz

DC iiBA

dx sin

)( Evaluate :Ex22

2122222221 2

1sinIIdz

iziziz

22)Res(2

.at pole one only and , term the to due

plane-halfupper the on choosed iscontour the ,for (1)

222222

R -R -

iiiziz

izixix

eeedxx

)()2()

at pole one only , term

the by plane-halflower the choose we ,for (2)

4)Res(

4)Res()(

0 and 0 As

222222

cos)(222

1sin)(

0 ,0 As

222222

iiiiii

eeeeee

eeedxx

Chapter 20 Complex variables. 20.2 Cauchy-Riemann relation A function f(z)=u(x,y)+iv(x,y) is...

Documents

Convergence of cauchy-riemann sums to cauchy-riemann integrals · Convergence of Cauchy-Riemann Sums to Cauchy-Riemann Integrals1 Otto Szcisz and John Todd Two general theorems giving

Training & Conditioning 20.2

Lagrange + Cauchy + Taylor

20.2. Electric Potential

Competencies Needed 20.2

4. Complex integration: Cauchy integral theorem and Cauchy

RenderNet: A deep convolutional network for differentiable ... · makes these techniques non-differentiable. Although it is possible to treat them as non-differentiable renderers

cauchy inequalities.ppt

NON-DIFFERENTIABLE FUNCTIONS

Chapter 20.2

15 Cauchy Gourst Thrm & Cauchy integral formula

Cauchy Fluxes and Gauss–Green Formulas for Divergence ...torresm/pubs/PublishedVersion_201… · 18/04/2019 · is, unbounded weakly differentiable vector ﬁelds in Lp whose distributional

Partial Diﬀerential Equations (Week 3+4) Cauchy-Kovalevskaya … · Cauchy-Kovalevskaya and Holmgren GustavHolzegel February1,2019 1 Introduction Recall the Cauchy-Kovalevskaya

Section 20.2-1

€¦ · Web viewCauchy distribution. Cauchy sequence, Cauchy stress tensor, Cauchy principal value, Cauchy's integral formulaAugustin-louis cauchY“I .am a Christian, that is to

(a) Write down the Cauchy-Riemann conditions which are necessary for a complex function to be differentiable at a point. (b) At what points is the function f(z) = differentiable. (c)

Cauchy Integral Formula

ELECTRIC CIRCUITS 20.2

Cauchy Euler

Calculus Review Session - Duke University · Differentiable functions • Differentiable: A function is differentiable at a point when there's a defined derivative at that point