ee2012 complex

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omp ex na ys somp ex na ys somp ex na ys somp ex na ys s

EE2012 ~ Page 9 / Part 2 ben m chen, nus ece

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FlowChartofMaterialinComplexAnalysisFlowChartofMaterialinComplexAnalysisFlowChartofMaterialinComplexAnalysisFlowChartofMaterialinComplexAnalysisz x iy ( ) ( ) ( )z t x t iy t ( ) ( , ) ( , )f z u iv u x y iv x y

Complex function of a complex

variable

~ define a complex mapping from

Complex function of

a real variable

~ define a curvecurve on

one comp ex p ane to anot era complex plane

( ) ( ) ( )C

f z dz f z t z t dt

( ) 0Cf z dzComplex integral

~ integration of a complex

function (of a complex variable)

Cauchys Integral theorem

~ valid whenf(z) is analyticalanalyticalfor every point encircled by a

~ upperboundofacomplexupperboundofacomplex

integrationintegration

~~ CauchyCauchy--Riemannequations,Riemannequations,

singularitiessingularities


to

be

continued

on

the

next

pageto

be

continued

on

the

next

page

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FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)( )f z

( )

0( )( ) n

f zf z

Another integral formulaCauchys integral formula

0Cz z 10 !( )

n

C nz z

~ n 0, Cis a closed curve enclosesz0andf(z) is analytic inside C

~powerseriesofananalyticfunctionpowerseriesofananalyticfunction

~ Cis a closed curve encloses

z0andf(z) is analytic for

every point inside C

A licationsofcom lexA licationsofcom lex

complex integral in terms of residues

0,C

z z z esintegralintegral

~ TaylorseriesTaylorseries,,LaurentseriesLaurentseries,,orderofsingularities,residuesorderofsingularities,residues

~ The most general result for


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FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)FlowChartofMaterialinComplexAnalysis(cont.)

A licationsofcom lexA licationsofcom lexintegralintegral

2

cos sin d

( )P x

x dx dx

cosf x mxdx

0 ( )Q x

sinf x mxdx

LaplacetransformLaplacetransform argumentprincipleargumentprinciple RouchestheoremRouchestheorem

MissioncompletedMissioncompleted


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Basic Operations of Complex NumbersBasic Operations of Complex NumbersBasic Operations of Complex NumbersBasic Operations of Complex Numbers

Cartesian and Polar Coordinates:Cartesian and Polar Coordinates:

1tanar 2 2

yii z x

Eulers Formula:Eulers Formula: cos( ) sin( )ie i

Additions:Additions: It is easy to do additions (subtractions) in Cartesian coordinate, i.e.,

( ) ( ) ( ) ( )a ib v iw a v i b w

Multi l ication:Multi l ication: It is eas to do multi lication division in Polar coordinate, i.e.,

( )( )i i ire ue ru e ( )i

i

i

re re


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Complex Functions of a Real VariableComplex Functions of a Real VariableComplex Functions of a Real VariableComplex Functions of a Real Variable

Complex functions of a real variable are needed to represent paths

or contours in the complex plane.

( ) ( ) ( ) , [ , ]z t x t i y t t a b

Example 1

5 0 2itz t e t 34

5

5cos 5sint i t

- 5 - 4 - 3 - 2 - 1 1 2 3 4 5-1

1

2

( ) 5cos( ) 5sin , [0,2 ]

x t t

y t t t

-4

-3

-2


-5

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Real Variable (cont.)Real Variable (cont.)Real Variable (cont.)Real Variable (cont.)

Normal differentiation and integration rules are applicable:

1 1 2 2 1 1 2 2( )c z c z c z c z

b b b

c c dt c dt c dt a a a

b

a


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CurvesCurvesCurvesCurves

The set of images Smooth closed curveSmooth closed curve

{ ( ) | [ , ]}C z t t a b 2

3

4Smoot h curve

- 2 2

2

Smooth curveSmooth curve

the complex plane3

- 4 - 2 2 4 - 2

The curve is smooth if

-

i ecewi se smooth curve

Piecewise smooth (pws) curvePiecewise smooth (pws) curve

is continuous( )z t


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Curves (cont.)Curves (cont.)Curves (cont.)Curves (cont.)

The length of a curvelength of a curve is

b

m

( )aL z t dt

A curve is thus a mapping

of the real number line

a b

onto the complex plane


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Two Special CurvesTwo Special CurvesTwo Special CurvesTwo Special Curves

CircleCircle

e parametr c escr pt on

for a circlecircle centred at r

complex point a and with aradius ris Re

a

( ) , [0,2 ]

it

z t a re t


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Two Special CurvesTwo Special CurvesTwo Special CurvesTwo Special Curves

Straight LineStraight Line

e parametr c escr pt on

of a straight linestraight line segmentb

with starting point a andendpoint b is

R ea

( ) ( ) , [0,1]z t b a t a t


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and Length of Curvesand Length of Curvesand Length of Curvesand Length of Curves

a) The line segment that connects the points and2 3i 5 6i( ) (7 9) ( 2 3) , [0,1]z t i t i t

1 1b

9

b The circle with radius 2 and centre 1 i

0 0a

z t t t t

7

( ) (1 ) 2 , [0, 2 ]itz t i e t

2 2 2

0 0 0

( ) 2 2 2 1 4b

it it

a

L z t dt i e dt i e dt dt


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Example 2 (cont.)Example 2 (cont.)Example 2 (cont.)Example 2 (cont.)

2, 2 3y x x

2 3x t t

c)

Let 2t

( ) ( 2), 2 3z t t i t t

3 3

( ) 1 2 2

b

L z t dt i dt dt 2 2a


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Complex Functions of a Complex VariableComplex Functions of a Complex VariableComplex Functions of a Complex VariableComplex Functions of a Complex Variable

A complex function of a complex variable maps one plane to another plane.

Im[w]Im[z]

w= f(z)

Re[z] Re[w]z0

0

Cimage

domain

( )f z wThese functions are of the form ( ) ( , ) ( , ) .f x iy u iv u x y iv x y


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Complex MappingComplex MappingComplex MappingComplex Mapping

A complex-valued function

defines a mappingmapping of a domain,D, onto its imageimage thew-plane.

, , ,

For any pointz0

inD, we call the point the image ofz0. Similarly,

the points of a curve Care mapped onto a curve on the w-plane.0 0

( )w f z

Im[w]Im[z]

w= f(z)

Re[z] Re[w]

z0w0


imagedomain

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Example:Example:zz22Example:Example:zz22

a) 2 2

( ) ( )w f z z x iy 2x y i xy

u iv

The functionf(z) =z2

would for example map the complex number 1 + i to i2and the number 2 i to 3 i 4.

The line segment or is mapped to

the line segment , since

[0, 2], 0x y , [0,2], 0x t t y

2 0 2 0u t t v 2 2 2( ) ( 0)u iv x iy t i t


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Example:Example:zz22 (cont.)(cont.)Example:Example:zz22 (cont.)(cont.)

2 2 2 2( ) ( )i i iw f z R e z r e r e In polar coordinate:

or examp e, e mage o e reg on

under the mapping is

,r

2w z 1 9 / 4, / 3 2 / 3R

v

2

x u


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Example:Example: eezzExample:Example: eezz

b zw z e x i y

e

cos sinx xe y i e y u iv

(1 ) cos1 sin1f i e i e For example,


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Example:Example: eezz (cont.)(cont.)Example:Example: eezz (cont.)(cont.)

In terms of polar coordinates

( ) i z x i y x i yw f z R e e e e e

Therefore

,R e y

Note that 0zw e z


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For , consider the images of:z

w e y v

1. Straight lines

and

0 constx x

0 consty y u

x0

x

x0e

From , we see,xR e y y v

circle and

0

0xw e 0y y

u

y0

x

y0

0arg( )w y


,x

R e y

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x


D / ,z x iy a x b 2. Rectangle { }:c y d

,

rom a , we can conc u e a any rec ang e w s e para e o e

coordinate axes is mapped onto a region bounded by portions of rays

an c rc es. ere ore e mage o s

D / ,i a bw Re e R e c d { }

y v

de d

ea


xba u

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3. The fundamental region :

-

y

,

origin. The strip is mapped onto the upper half-plane0 y

-1

x u

1

More generally, every horizontal strip is mapped onto the

-

2c y c


.

,

x

R e y

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More ExampleMore ExampleMore ExampleMore Example

c) ( )

iz ize e

f z

( ) ( )

2 2

i x i y i x i y ix y ix ye e e e

(cos sin ) (cos sin )2

y y

e x i x e x i x

cos sin2 2

cos cosh sin sinh

e e e ex i x

x y i x y

u iv


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More ExampleMore ExampleMore ExampleMore Example

Alternatively, ( ) cos cos( )

cos( )cos( ) sin( )sin( )

f z z x iy

x iy x iy

cos( )cosh( ) sin( )sinh( )x y i x yu iv

sin( ) sinh( ), cos( ) cosh( ), tan( ) tanh( ).ix i x ix x ix i x since

Keep in mind that

sin cos2 2

iz iz iz ize e e e

z zi

sinh cosh2 2

z z z ze e e e

z z


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ObservationsObservationsObservationsObservations

Identities for trigonometric functions also hold for complex arguments, e.g.

2 2

1 2 1 2 1 2

cos sin 1

sin( ) sin cos cos sin

z z

z z z z z z

1 2 1 2 1 2cos( ) cos cos sin sinsin( ) cos , cos( ) sin

z z z z z zz z z z

Also note that

s n s n cos cos s n ,

cos cos cosh sin sinh

z x y x y

z x y i x y


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ObservationsObservationsObservationsObservations

is continuous at the point if and are( )f z w ( , )u x y ( , )v x y0 0z x i y

.

For example, is continuous everywhere, because

0 0

2( )f z z 2 2u x y

and are continuous everywhere.

It can be shown that the re ular rules of differentiation and inte ration

2v xy

are still valid, e.g.

sin cosd z zdz

1n ndz n zdz


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Complex IntegralComplex IntegralComplex IntegralComplex Integral

Consider the curve and the complex function f which: ( ) , [ , ]C t z t t

is continuous on C. The complex integral of f along Cis then defined as

( ) ( ) ( )C

f z dz f z t z t dt


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Example 4Example 4Example 4Example 4

a) Let b) Let Cbe a circle with radius r: ( ) 2 3 , 1 2C z t t i t t

and2( )f z z

2( ) 1/f z z

1

2

2 3 2 3C

z z t i t i t : ( ) , [0,2 ]

it

C z t re t

1

3 7

(2 3)i t dt

2

0

1/it

it

C

irez dz dt

re

3

3223 21i

2

0

2i dt i


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Properties of Complex IntegralsProperties of Complex IntegralsProperties of Complex IntegralsProperties of Complex Integrals

As for real integrals, the following rules apply:

1. [ ( ) ( )] ( ) ( )f z g z dz f z dz g z dz C C C

2. ( ) ( ) , complexC C

k f z dz k f z dz k


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Properties of Complex IntegralsProperties of Complex IntegralsProperties of Complex IntegralsProperties of Complex Integrals

3. 4.( ) ( ) ( )f z dz f z dz f z dz ( ) ( )f z dz f z dz 1 2 *C C

C*C C2

CC1


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Estimation of a Complex IntegralEstimation of a Complex IntegralEstimation of a Complex IntegralEstimation of a Complex Integral

Letf (z) be continuous on . If on C, then: ( ) , [ , ]C t z t t ( )f z M

( )f z dz M LC

, . .

2 2 ( )L z t dt dtdt dt


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Estimation of Complex IntegralEstimation of Complex Integral An IllustrationAn IllustrationEstimation of Complex IntegralEstimation of Complex Integral An IllustrationAn Illustration

Graphically, take real integration as an example,

f(t)

M

t

L = b a

( )b

a

f t dt L shaded area with red lines


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Estimation of Complex IntegralEstimation of Complex Integral An IllustrationAn IllustrationEstimation of Complex IntegralEstimation of Complex Integral An IllustrationAn Illustration

For complex cases, for example, we takef(z) =z3 and Cto be a unit circle

1

0.8

1

( )f z z ( )

1

f z z

M

-0.5

0

.5

0.4

0.6

.

0.5

1

0

0.5

1

-1

0.5

1

0

0.5

1

0

.

-1

-0.5

0

-1

-0.5

-1

-0.5

-1

-0.5

( ) 2f z dz M L


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Example 5Example 5Example 5Example 5 4 6i

Find the upper bound for the absolute value of z

C

e dz

2 3i L

w ere s e ne connec ng e po n s an2 3i 4 6i

( ) (2 3) (2 3), [0,1]z t i t i t 1

2 22 3 13L dt , ,

(( ))

t i t

x ti yt

t

0

60 M

z x y x y x y xe e e e e e e

40

50

( )z te

2 2 4

54.5982x t

e e eM

Thus,413 196.8566ze dz M L e 10

20

largest 1x t


C

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

t

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Example 6Example 6Example 6Example 6 a b a b

Show that2

| | 2

1 4

31zdz

z

4

6 f(z)

Noting that-4

-2

0

=

2 2 2 2

( 1) ( 1) 1 1 1 1z z z z -2

-1

0

12

-2

-1

0

1

2

-6

L

0.3

0.32

0.34

,

2 2 1 1

1 1 3z z M ( ) itf z

M

0.24

0.26

0.28

2| | 2

1 4

31z

z

dz MLz


0 1 2 3 4 5 6

0.2

0.22

t

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Analytical FunctionsAnalytical FunctionsAnalytical FunctionsAnalytical Functions

A functionf (z) defined in domainD is said to be an analytic function

points ofD. The functionf (z) is said to be analytic at a pointz0 inD if

z0.

nei hborhood onnei hborhood on

D z0

a complex planea complex plane

is a discis a disc


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Discrepancy of Definitions of Analytic FunctionsDiscrepancy of Definitions of Analytic Functions

e n on g ven y mos popu ar ex s e.g., e re erence ex :e n on g ven y mos popu ar ex s e.g., e re erence ex :

A functionf (z) defined in domainD is said to be analytic if it is differentiable

at every point ofD.

Definition given in some odd books or the text by Garg et al.:Definition given in some odd books or the text by Garg et al.:

A functionf (z) defined in domainD is said to be analytic if it is differentiable

with a continuous first order derivative at ever all oint ofD.

Nevertheless, we will use the definition given by Garg et al. throughout


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SingularitiesSingularitiesSingularitiesSingularities

Points where a function is not analytic are called singular pointssingular points or

.

Example:

is analytic everywhere

inD exceptz = 0, which is thus X

1

( )f z z

the singular point or pole of the

.

Note that a function is either analytic or singular at any given pointNote that a function is either analytic or singular at any given point


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Observations:Observations:

It can be shown that the existence of continuous first derivatives implies

the existence of a continuous second derivative, etc. This also implies

e ex s ence o a ay or ser es.

20( )z z

An analytic function can thus also be defined as a function for which a

2!

Taylor series expansion exists.


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TheoremTheoremTheoremTheorem

If a functionf (z) = u + i v is analytic inD, then the following Cauchy-Riemann

, . .

andx y y x

Alternatively, if the Cauchy-Riemann equations hold for a functionf (z) = u + i v

,

analytic inD.


Proof.Proof.Proof.Proof.

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( )f z


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ExampleExampleExampleExample

2 2 2

u v u v ,x y y x

an t e part a er vat ves are cont nuous .z

, .


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*

f z z x iy u iv

u v u v , ,

x y y x

( )f z is not analytic anywhere


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2 2 2 2( ) 2f z x y i x y u iv

2 2 2 22 , 4 , 2 , 4u v u v

xy x y x y xyx x

The Cauchy-Riemann equations only

hold forx = 0 and/ory = 0. Since the

function is not analytic in ay = 0 x = 0

neighbourhood ofx = 0 ory = 0,f (z) is

not analytic anywhere.


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Cauchys Integral TheoremCauchys Integral TheoremCauchys Integral TheoremCauchys Integral Theorem

A domainD in the complex plane is called simply connectedsimply connected if every closed

.

called multiply connectedmultiply connected.

Simply connected Doubly connected Triply connected


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Cauchys Integral TheoremCauchys Integral TheoremCauchys Integral TheoremCauchys Integral Theorem

Iff (z) is analytic in a simply connected domainD, then for every closed path

CinD

( ) 0C

f z dz Fundamental Theorem of CalculusFundamental Theorem of Calculus

D

If F(z) is an analytic function with a

continuous derivativef (z) = F'(z) in a

Creg onD con a n ng a p ecew se

smooth (pws) arc :z =z(t), t

( ) ( ( )) ( ( ))f z dz F z F z

z


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Applications of Cauchys TheoremApplications of Cauchys TheoremApplications of Cauchys TheoremApplications of Cauchys Theorem

ApplicationsApplications:: z2

1. Iff (z) is analytic in a simply connected

domainD, then the integral off (z) isC1

*C2

independent of path inD.

2

z1*

1 2

( ) ( ) 0C C

f z dz f z dz

*1 22

( ) ( ) ( )C CC

f z dz f z dz f z dz


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Applications (cont.)Applications (cont.)Applications (cont.)Applications (cont.)

3. The integral along a closed path C1 of the functionf (z) which is analytic

in the multiply connected domainD, is given by the sum of the integrals

around paths which encircle all openings within the region bounded by

C1, e.g.

C1* *( ) ( ) ( ) 0

C C C

f z dz f z dz f z dz

*C2 *C3

* *

Thus ( ) ( ) ( )f z dz f z dz f z dz C2 C3

1 2 3

2 3

( ) ( )C C

f z dz f z dz


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Applications (cont.)Applications (cont.)Applications (cont.)Applications (cont.)

4. In general, it can be shown that if the path Cencloses the pointz0, then

0( )2 1

n

C

z z dzi n

Proof. Without loss of any generality, we assumez0 = 0. For n 0,zn is

analytic anywhere on the whole complex plane. By Cauchys theorem, its

integration over any closed curve is 0. For n = 1, By Application 2, the

integration over any path enclosedz0 is the same. It was shown earlier

: ( ) , [0,2 ]tC z t re t a e n egra on o z over a c rc e

2 2

1itire


0 0it

Cz re

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Proof of Application 4 (cont.)Proof of Application 4 (cont.)Proof of Application 4 (cont.)Proof of Application 4 (cont.)

For n < 1, let m = n > 1 and integrate over . We have: ( ) , [0,2 ]itC z t re t

( 1)1

0 0

1 i m tm m imt m

C

ire idz dt e dt

z r e r

1

0

2( 1)

( 1)1

m t

m

i m

e

i mr

1

1

( 1)

1cos 2( 1) sin 2( 1) 1

m

m

m r

m i m

11

1 0 1 01 m

m r

m r


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(a) with Cthe ellipse1/

C

z dz 2 24 1x y

The equation of the ellipse is given by2 2

0 02 2

( ) ( )1

x x y y

a b

2 a

(x 0 , y 0)2 b


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Example (cont.)Example (cont.)Example (cont.)Example (cont.)

Consequently, the ellipse is centred at the origin. The function

= =

2 24 1x y

.

is therefore the same for any path which encloses the origin. Cmay thus

be re laced with a circular ath of radius 1 i.e. it , ,

1

C2

12

it

it

iedz dt i

z e


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22

2

09

z

z

edz

z

The function is2( ) /( 9)zf z e z

analytic inside the region enclosed

by the curve.

-3 3


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1 1 1/ 2 1/ 2

1 ( 1 )z z

z i z i

The function is analytic inside

1

the region enclosed by the path

-1


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Cauchys Integral FormulaCauchys Integral FormulaCauchys Integral FormulaCauchys Integral Formula

0 , [0, 2 ]it

z z Re t 0zConsider a circle with centre . Then

2

0

0

( )itdt i f z Re dt

0

( )

C

f zdz

z z

2

0

0

( )it itit

f z ReiR e

Re

Sincef (z) is continuous and the integral will have the same value for all

values ofR, it follows that

2 2 2

0 0 0 00

0 0 0 0

lim ( ) ( ) ( ) 2 ( )R

C

dz i f z Re dt i f z dt i f z dt i f zz z


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Cauchys Integral FormulaCauchys Integral FormulaCauchys Integral FormulaCauchys Integral Formula

This leads to Cauchy's integral formulaCauchy's integral formula, stating the following:

Letf (z) be analytic inD. Let Cbe a closed curve inD which encloses

z0. en

00

( )2 ( )

C

f zdz i f z

z z


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Power Series as Analytic FunctionPower Series as Analytic FunctionPower Series as Analytic FunctionPower Series as Analytic Function

A power series is of the form

2 0( )c z z

n

0 0 1 00

( ) ( )nn

c z z c c z z

convergence r* such that the series is absolutely convergent for

00

n

n

c z z

and divergent for .*0z z r *

0z z r

* =

21 1


1 z

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Power Series as Analytic FunctionsPower Series as Analytic FunctionsPower Series as Analytic FunctionsPower Series as Analytic Functions

The radius of convergence is

r*

z0

Convergent

*

1

lim nn

n

r

c

Divergent Each power series defines a

the radius of convergence.


Example:Example: 1

1

z zz

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Power SeriesPower Series

Power SeriesPower Series

Inside the radius of convergence, the power series

00

( ) ( )nnn

f z c z z

2 2z z

can be integrated and differentiated on a term-by-term basis, i.e.

1 1

00

( ) ( )nnnz z

f z dz c z z dz

1 *0 0

0

( ) ,nnn

nc z z z z r

( )d f zdz


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ExamplesExamplesExamplesExamples

3(a)4 3 0

1

sin3! z

z

dz zz dz

6

z 3

42( 1)

z

dzz

3 13!

z

zz e

dz 13

3 3ze ze

(b)


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of Complex Functionsof Complex Functionsof Complex Functionsof Complex Functions

If a functionf (z) is analytic for 0 ,z z R

thenf (z) has a Taylor seriesTaylor series expansion

R

where

00

n

n

z a z z

z0

10

1 ( )

2 ( )n n

C

f za dz

i z z

( )0( )

!

nf z

n


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of Complex Functionsof Complex Functionsof Complex Functionsof Complex Functions

If a functionf(z) is analytic in

,thenf (z) has a Laurent seriesLaurent series

ex ansion

1 0 22

z0R1

C

nz a z z

where

n

10

1 ( )

2 ( )n n

C

f za dz

i z z

Points where a function is not analytic

are called singularities.


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Example 10Example 10Example 10Example 10z

e sinz cosz(a) The functions , and are analytic functions, and have

0

2 3

1z z z

e z *1 nc

3 5 7

2! 3!

z z z

1!

1

nn

nn c

s n3! 5! 7!

z z m1

( 1)!

n

n

cos 12! 4! 6!

z z zz lim!

lim( 1)

n

nn

n


n

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(c) The function is analytic for . The Taylor series expansion with1

1 z1z

z0 = , . .,

21 1 z z

(d)* Find Taylor series expansion of atz0 = 3 & its convergence radius.

1 z

1( )f zz

2 3

2

2 3

3 4

11 1 3 3 33( ) 1

3 3 3 3

1 3 ( 3) ( 3)

3 3 3 33 z

zz z z zz

z

zf

The series converges for all Thus, its

3

| 3 .3

313

zz

* 3.r


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cos 1z

2 4 61

1 1

z z z

3 5z z z

a z 2! 4! 6!z 2! 4! 6!

e unc on z as a remova e a z0 = 0.

(b) 5sin

( ) z

f zz

3 5 7

5

1

3! 5! 7!

z z zz

z

2

4 2

1 1 1

5! 7!3!

z

z z

The functionf (z) has a 4th order pole atz0 = 0.


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2 1/

( )

z

f z z e2 1 1 1

1z

(c) 2 1 1

z z 2! 3!z z z

The function z has an essential sin ularit atz = 0.

2! 3!z

2 2 2 (d)

2 2 2( )

( 1) ( 1) ( 1)

1 2

f zz z z

Thus, z has 2nd order ole atz = 1.

2 ( 1)( 1) zz


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Forz0 = 1, we have the following Laurent series off(z) centered atz0 = 1

1 ( 1)( ) 1

( 1) 1

1

11

z

zf z

z z zzz

2 31

1

1 3 2( 1) 2( 1) 2( 1)

z z zz

z z z

Thus, the order of singularity off (z) atz0 = 1 is again equal to 1.

1st order poles are also called simple poles or simple singularities.


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ZerosZerosZerosZeros

If , then the function g (z) is said to have a zero or root in z =z0.0( ) 0g z

If and , then the function( )

0( ) 0ng z ( 1)0 0 0 0( ) ( ) ( ) ( ) 0

ng z g z g z g z

n z =z0.

If the function g (z) has an nth order zero inz =z0, then has an1

( )f z

n-th order pole inz =z0.


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a) Consider the function , which has a singularity at1

( )( 1)( )z

f zz e e

0 1z

Let ( ) ( 1)( )zg z z e e

Then ( ) ( 1) (1) 0z z

g z e e z e g

Therefore has a 2nd order zero in = 1 and has a 2nd order

( ) ( 1) (1) 2 0z z zg z e z e e g e

pole inz0 = 1.


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ExampleExampleExampleExampleAn alternative solution to Q.2.2.1 in Tutorial 2.2, i.e., finding the order of

oles for

2

( ) sin 1( )

( )

zh z e z

f zg z z

1. It is obvious that g (z) has a zero of order 2 atz = 0, i.e., n = 2.

2. Noting , we have m = 2.0

0(0) cos 0

z

z

z

e z

h e z

3. The order of the pole off(z) atz0 = 0 is given n m = 0. It is removable.

0s n ze z


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ResiduesResiduesResiduesResidues

We know that iff (z) is analytic in domainD except at the pointz0, it has a

n

Laurent series expansion of

n

n

with

10

1 ( )

2 ( )n n

C

f za dz

i z z

and Cany closed curve inD which enclosesz0.


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idididid

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ResiduesResiduesResiduesResidues1 2, , nz z zIff (z) is not analytic in several points , then

1

( ) ( )C C

f z dz f z dz 2

( ) ( )

nC C

f z dz f z dz

1 2es , es , es ,

2 Res( , )

n

n

i

z z z

i f z

C

C

1i

C1 z2

z3C3

z1


C l l ti f R idC l l ti f R idC l l ti f R idC l l ti f R id

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Calculation of ResiduesCalculation of ResiduesCalculation of ResiduesCalculation of Residues

1. f(z) has a simple pole atz =z0:0

0 0Res( , ) lim ( ) ( )z z

f z z z f z

2. f (z) has an nth order pole atz =z0:

0

1

0 01lim1

Res( , ) ( ) ( )( 1)!z z

nn

n

df z z z f zn d z

3. where has a simple zero at , while0z z( )

( )( )

A zf z

B z

0( ) 0A z ( )B z

and bothA andB are differentiable atz =z0: 000

( )Res( , )

( )

A zf z

B z


C l l i f R idC l l i f R idC l l i f R idC l l i f R id

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Calculation of ResiduesCalculation of ResiduesCalculation of ResiduesCalculation of Residues

2'. The 2nd formula of the previous page can be modified as follows:

- = ,

11 md

00 01

es , m( 1)! mz z

z z z zdm z

where m is any integer with mn.

This formula was proposed and derived by Phang Swee King, a

student taking EE2012 in Semester 2 of Year 2007/08. It may yield

a simpler way in computing the residue for certain situations.


E l i F l 2E l i F l 2''E l i F l 2E l i F l 2''

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Example using Formula 2Example using Formula 2''Example using Formula 2Example using Formula 2''Example: The extra freedom in selecting m in Formula 2' can simplify

some problems in computing residues and thus complex integrals. We

consider the following example (which was solved earlier),

sinz

It was shown that the function has a 3rd order pole atz0 = 0. However, if

4

| | 1z

zz

we use m = 4 instead of 3, it is much easier to compute the associated

residue compared with that using the original formula 2, i.e.,

3 3

4

4 3 4 30 0| | 1

sin 1 sin 1 12 lim 2 lim sin 2 cos 0

3! 3! 3! 3z zz

z d z d idz i z i z i

z dz z dz


E l 13E l 13E l 13E l 13

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22

4 3.

z

zdz

z z

(a) Calculate 00 0

Res( , ) lim ( ) ( )z z

f z z z f z

2

4 3 4 3( )

1

z zf z

z zz z

has simple poles inz0 = 0 andz0 = 1.

4 32 Res 0 Res 1

zdz i

C

2

4 3 4 32 lim lim

z z z

z zi

1 200 1

2 [ 4 1] 6

z zz z

i i


E l 13E l 13E l 13E l 13

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(b) Compute

C1z

e dzz

1

2 Res( ,0)z

z

edz i f

z

100

2 lim zz

i e


00 0,

z zz z z z

E l 13E l 13E l 13E l 13

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2

sinzdz(c) Compute

1z

sinz has a sim le ole at = 0.

C

z

Thus 10

sin

2

sin sin2 lim 2

zzz zdz dz i i

1 1z z


00 0,

z zz z z z

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2

2 4

6

zdz

(f) Compute C.z

2 4z 2 4z 2.5-3 22 6

zz z

( 3)( 2)z z

as a s mp e po e a z0 = enc ose y . us,

2 4 2 4 16z z i 2 2

2.5 , 3 56 zz zz z


00 0Res( , ) lim ( ) ( )

z zf z z z f z

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C(a) Evaluate

21

d

21 1 1

1-2+ 3-2- 3

0

1 120 1

2

cos 2 2

12

zz z i z

i dz

1

1

12

2 3 2 3

z

i dzz z

22 2 Res( , 2 3)

3i i f


00 0Res( , ) lim ( ) ( )

z zf z z z f z

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FunctionsFunctions

FunctionsFunctions

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FunctionsFunctionsFunctionsFunctionsWe now consider real integrals for which the interval of integration is not

finite. These are called improper integrals, and are defined by

( ) lim ( )R

f x dx f x dx

Assume that the is a real rational function with

R

( )( ) P xf x

for all realx (i.e. no real poles)( ) 0Q x

degree[ ( )] degree[ ] 2( )Q x P x


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S

Improper IntegralsImproper IntegralsImproper IntegralsImproper Integrals-R R

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R

Then

( ) ( )R

f z dz f z dz ( )f z dz (*)

The 2nd condition, i.e., degree [Q(x)] degree [P(x)] + 2, implies if

C R S

1 11 1 1 1( ) , ( )

n mn mn n m mQ z q z q z q P z p z p z p

122 1

2( ) 1( ) ( )

mm

mm mp pz p zz P z K K Kz f z K f z

, . ,

1 nn nz z q z zqq zz

( ) 0 asK K

f z dz M L R R

(Result S)(Result S)


S

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2 3

1

1dx

x

(b) Calculate +i C

2 3

1( )f z

3 3

1

Let

-i

2 3

12 Res( , )

(1 )dx i f i

x

5

12 6 3lim

16 8( )z i

ii i

z i

11 nd


00 01

mes ,( 1)!z z n

z z z zn d z


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4

1

4dx

x

(c) Calculate . Let and its poles are given by41

( )4

f zz

(2 1)14 4 (2 1) 444 0 4 4 4

i ni n

z z e z e

4

(2 0 1)4 4

2 , 0, 1, 2, 3,n

i i

z e n

C1

(2 1 1) 34 4

2

,

2 2 1 , 1i i

z e e i n

-1+i 1+i

54 4

3

(2 3 1) 74 4

2 2 1 , 2i i

i i

z e e i n


4 ,


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Then

41

2 Res( ,1 ) Res( , 1 )4

dx i f i f ix

-C

3 31 1

1 12

4 4z i z ii

z z

1 1

2 (1 ) (1 )16 16

i i i

2 8 4i

0( )A z


00

,( )B z

Improper Integrals of FourierImproper Integrals of Fourier--typetypeImproper Integrals of FourierImproper Integrals of Fourier--typetype

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Consider integrals of the form

or

mxdxxf cos

mxdxxf sin

Assume thatf (x) = P(x) / Q(x) is a real rational function with

Q(x) 0 for all realx (i.e. no real poles)

m > 0


Improper Integrals of FourierImproper Integrals of Fourier--typetypeImproper Integrals of FourierImproper Integrals of Fourier--typetype

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Consider the complex integral

Rimz imz imz imz

We will show (i.e., Theorem X next page) that under the conditionsm > 0 and

,jjC R S

egree [Q x ] egree [P x ] + 1, we ave . o ng

that , it can be showncos sinimz

e mz i mz

( ) 0 asmz

S

f z e dz R

( )cos Re 2 Res( , )imz jj

f x mx dx i f e a

S

( )sin Im 2 Res( , )imz jf x mx dx i f e a

-R R


Theorem XTheorem XTheorem XTheorem X

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If with and m > 0, then( )

( ) P z

g z degree[ ( )] degree[ ( )] 1Q x P x

S

lim ( ) 0mzR

S

g z e dz

-R R

There is no name for this theorem. As such, for easy references, we call it

Theorem X. The result has been used earlier in deriving improper integrals

of Fourier-type. It will be used later few more times.


Proof of Theorem XProof of Theorem X (self(self--study)study)Proof of Theorem XProof of Theorem X (self(self--study)study)

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S

iYObserving the curves on the right, if we let

X Y andR , the inte ral

of the function along Sis the same as it

along the blue straight lines. Under that

-R RX Xdegree [Q(x)] degree [P(x)] + 1

,

( ) ( ) ,| | | |

K K Kzg z K g z

z X iy X

( )| | | | | |imz im X iy imX my mye e e e e

and thus

1

( ) 1 0 asX iY Y

imz my mY K K Kg z e dz e dy e X


0 0X i

Proof of Theorem X (cont.)Proof of Theorem X (cont.) (self(self--study)study)Proof of Theorem X (cont.)Proof of Theorem X (cont.) (self(self--study)study)

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S

iYSimilarly, we can show the integral along

the strai ht line from X+ iYto X+ i0 has

a same bound, i.e.,

0X i

-R RX X

The integral along the line fromX+ iYto X+ iY, we have

( ) 0 asimz

X iY

g z e dz X mX

( ) ( ) ,| | | |

K K Kzg z K g z

z x iY Y

( )| | | | | |imz im x iY imx mY mY e e e e e

and thus

( ) 2 0 as ,X iY XmY

imz mY Ke Xg z e dz dx K e X Y


X iY X QEDQED

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(b) 2 2sin Im 2 Res ,

1 1

izx x z edx i ix z

+i C

Im 2

2

z

z i

zei

z

Im

/

e

e

-i

0( )A z


00

,( )B z


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2i z

c 2 2Re 2 Res ,39 9dx i ix z +i

C3i

3

Re 2

2 z i

ei

z

6 6

Re3 3

e e

-i3i

0( )A z


00

,( )B z

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Simple Poles on the Real AxisSimple Poles on the Real AxisSimple Poles on the Real AxisSimple Poles on the Real Axis

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Since g (z) is differentiable, it is continuous and bounded, and therefore

( )

( )C

g z dz M L

0( ) 0M C()

This results in Jordan's lemma, stating that

a

lim ( )f z dz 1 Res( , )i a i f a

( )C



i

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sinxdx

x(a) Calculate

Consider , with Cthe path as indicated in the illustration.

iz

C

edz

zBecause the function is analytic in

the domain enclosed by C, and according to

( ) e

f zz

Cauchy's integral theorem, the integral must

be zero, i.e.,S

i z

C

edz

z

( )

R

R C S

0

i zedz

z

R

-R -



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Let and . Then, by Jordans lemma0 R

0 0( )

lim ( ) Res( ,0)= lim izz

C

f z dz i f i e i

By Theorem X, we have .lim ( ) 0R S

f z dz

i z

edz

z

R

ize

dz

0

( )iz

edz i 0 0

ize

dz

0

ize

d i

sinx

cosx


z

xx


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Si 2

2

1

( 1)( 2)dx

x x

(b) Calculate C()

-

Consider

-

-1- -1+

-i 2

2

1dx

1 R 2

1dz 2 Res 2i i

C ( ) 1R C S


Exam le 17Exam le 17Exam le 17Exam le 17

Note

2 Res( , 2) 2 lim3

i f i i

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2( , )

3( 1)( 2) 2

1

z if

z z i

i

20 1( )

m es , = m 32

lim 0

zC

z z z

z dz

Result S

S

C

i 2

and

RS

R-R

-1- -1+

-1

2

1

( 1)( 2)C

dxz z

( ) 1R C S

21

( 1)( 2)dz

z z -i 2

2 Res( , 2)i f i 1 1 2 1i i

d dx


3 3( 1)( 2) ( 1)( 2)2 3 2z z x x

TransformsTransforms

TransformsTransforms yC

L l t f f f ti i d fi d

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Laplace transform of a functionf (t) is defined as

c x

0( ) ( ) ( )

st

F s L f t e f t dt

Inverse transform: 1( ) ( ) ( )2

st

C

f t L F s e F s dsi

Conditions for the existence of an inverse Laplace transform ofConditions for the existence of an inverse Laplace transform ofFF((ss):):

lim 0 lim finiand teF s s F s

In terms of the complex variablez:1

( ) ( ) Res[ ( ) , ]zt zt if t e F z dz F z e s

s s


iC


1

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Find the inverse Laplace transform of2 2

1( )F s

s

Solution:

zt ztC

y

2 2 2 2( ) Res , Res ,

zt zt

f t i iz z

z i z i

i t i t

z i z i

c

2i


00 0es , m

z zz z z z

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y

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Calculate cotz dz C

Note that with( )

cot f z

zz

( ) sinf z z

33 0 22 x

2 0 2 , .

Then, according to the argument theorem,

( )cot 2 (5 0) 10( )C C

f zz dz dz i if z


Application of Argument PrincipleApplication of Argument PrincipleApplication of Argument PrincipleApplication of Argument Principle

Let us focus on the case when f (z) is a rational function which is analytic

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Let us focus on the case whenf (z) is a rational function, which is analytic

except at possibly a finite number of points. The argument principle implies

1 ( ) 1 1

ln ( ) ln ( ) arg[ ( )]f z

dz d f z d f z i f z

2 2

1 ln ( ) arg[ ( )]

C C C

z z

d f z d i f z

2

2

1 1

1ar

10 ar

z z

z

z

i z z n

1

1

22

z

zi

. 1z Note that the ln is a real function.EE2012 ~ Page 138 / Part 2 ben m chen, nus ece

21 arg[ ( )]

z

f z n p

Case 1: If there is one zero and no pole of f ( ) encircled1z

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Case 1: If there is one zero and no pole off (z) encircled

- ,

2

1

arg[ ( )] 2 ( ) 2 (1 0) 2z

zf z n p

i.e.,f (C) will encircle the origin on the image plane once anti-clockwise.

C

C f(z)

. .


2z1z 1( )f z

2

21 ar [ ( ]

z

f z n p

Case 2: If there is no zero and one pole of f (z) encircled1z

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Case 2: If there is no zero and one pole off (z) encircled

- ,

2

1

arg[ ( )] 2 ( ) 2 (0 1) 2z

zf z n p

i.e.,f (C) will encircle the origin on the image plane once clockwise.

C

C f(z)

. .x


2z1z 2( )f z

1

21 ar [ ( ]

z

f z n p

Case 3: If there is no zero and no pole (or equal numbers1z

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Case 3: If there is no zero and no pole (or equal numbers

- ,

2

1

arg[ ( )] 2 ( ) 2 0 0z

zf z n p

i.e.,f (C) will not encircle the origin on the image plane at all.

C

f(C)f(z)

. .EE2012 ~ Page 141 / Part 2 ben m chen, nus ece

2z1z 1( )f z

2

''''

Let functionsf (z) and g (z) be analytic everywhere on and inside C. If

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( ) ( )f z g z

on C, then the total number of zeros ofp (z) =f (z) + g (z) inside Cis

.

C

f(z)

C


g (z)

Proof of Rouche's TheoremProof of Rouche's Theorem (self(self--study)study)Proof of Rouche's TheoremProof of Rouche's Theorem (self(self--study)study)

Let t [0, 1]. Sincef (z) and g (z) are analytic everywhere on and inside C

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and since on C, we havef (z) + t g (z) 0 for anyzC(why?).z z

Let

1 ( ) ( )f z tg z

=

2 ( ) ( )Ci f z tg z

i.e., (t) is integer-valued. It can also be shown (pretty complicated!) that (t)

is a continuous function of t. Thus, t can onl be a constant, which im lies

that the number of zeros off (z)+ t g (z) inside Cis constant for all t [0, 1].

The result of Rouches theorem follows by letting t = 0 and t= 1, respectively.


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Show that all the roots (zeros) of ,6 5 4 2( ) 0.1 0.2 0.3 0.1p z z az z z z

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where a is unknown, lie within the unit circle, i.e., , if a < 0.3.

Let 6 5 4 2( ) , ( ) 0.1 0.2 0.3 0.1f z z g z az z z z . Then on the unit circle:

5 4 2 5 4 2

6( ) 1f z z

. . . . . . . .

| | 0.1 0.2 0.3 0.1 | | 0.7 1a a

Thus, on Candf (z) and g (z) are analytic on and inside C. As

f (z) has 66 roots inside C,p (z) =f (z) + g (z) also has its 66 roots inside C.

( ) ( )f z g z



a = 0 295 a = 0 1 + i 0 28

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1

a = 0.295

1

a = 0.1 + i 0.28

0.5 0.5

0 0

-0.5 -0.5

-1 -0.5 0 0.5 1-1 -1 -0.5 0 0.5 1-1


ee2012 complex

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