Complex Variables Chapter 18 Integration in the Complex Plane

March 12, 2013 Lecturer: Shih-Yuan Chen

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Contents Contour integrals Cauchy-Goursat theorem Independence of path Cauchy’s integral formulas

2

Contour Integrals Recall that the def. of starts with a real function y = f (x) that is defined on [a, b]

on the x-axis. This can be generalized to integrals of real

functions of two variables defined on a curve C in the Cartesian plane.

( )∫b

adxxf

3

Contour Integrals For the integral of a complex fx f (z) that is

defined along a curve C in the complex plane. C could be smooth curve, piecewise smooth curve,

closed curve, or simple closed curve, and is called a contour or path.

C is defined in terms of parametric equations x = x(t), y = y(t), a ≤ t ≤ b, where t is a real parameter.

Using x(t) & y(t) as real & imaginary parts, we can describe C in the complex plane by means of a complex-valued function of a real variable t: z(t) = x(t) + iy(t), a ≤ t ≤ b. 4

Contour Integrals An integral of f (z) on C is denoted by or if C is closed. It is referred to as a contour integral or a

complex line integral.

( )∫C dzzf( )∫C dzzf

5

Contour Integrals f (z) = u(x, y) + iv(x, y) Let f be defined at all points on a smooth curve C

defined by x = x(t), y = y(t), a ≤ t ≤ b. Divide C into n subarcs according to the partition

a = t0 < t1 < … < tn = b. Corresponding points on C: z0 = x0 + iy0 = x(t0) + iy(t0), z1 = x1 + iy1 = x(t1) + iy(t1),

…, zn = xn + iyn = x(tn) + iy(tn). Let ∆zk = zk − zk−1, k =1, 2, …, n & let ||P|| be the

norm of the partition. Choose a point on each subarc. Form the sum

***kkk iyxz +=

( )∑ =∆

n

k kk zzf1

* 6

( ) ( ) ( )1 lim1

*

0∑∫

=→

∆=n

kkkPC

zzfdzzf

Contour Integrals Def. Let f be defined at points of a smooth

curve C defined by x = x(t), y = y(t), a ≤ t ≤ b. The contour integral of f along C is The limit exists if f is continuous at all points on C

& C is either smooth or piecewise smooth.

7

Contour Integrals Method of evaluation Write (1) in the abbreviated form

Thus, a contour integral is a combination of two real-line integrals.

( ) ( )( )( ) ( ){ }

( )2

lim

lim

∫∫∑∑

∑∫

++−=

∆+∆+∆−∆=

∆+∆+=

CC

C

dyudxvidyvdxu

yuxviyvxu

yixivudzzf

8

( )( ) ( )∫ ′b

adttztzf

Contour Integrals Since x = x(t), y = y(t), a ≤ t ≤ b, the RHS of (2) is

If we use z(t) = x(t) + iy(t) to describe C, the last result is the same as when separated into two integrals.

( ) ( )( ) ( ) ( ) ( )( ) ( )[ ]∫ ′−′b

adttytytxvtxtytxu ,,

( ) ( )( ) ( ) ( ) ( )( ) ( )[ ]∫ ′+′+b

adttytytxutxtytxvi ,,

9

( ) ( )( ) ( ) ( )3 ∫∫ ′=b

aCdttztzfdzzf

Contour Integrals Thm. If f is continuous on a smooth curve C

given by z(t) = x(t) + iy(t), a ≤ t ≤ b, then

If f is expressed in terms of z, then to evaluate f (z) we simply replace z by z(t). If not, we replace x & y wherever they appear by x(t) & y(t), respectively.

10

Contour Integrals Ex. Evaluate , where C is given by x = 3t, y = t2, −1 ≤ t ≤ 4. (sol)

∫C dzz

( ) ( )( )( )

( )( )

( ) idttidttt

dtitittdzz

ittitttzfittzitttz

C

65195 3 29

233

33233

4

1

24

1

3

4

1

2

22

2

+=++=

+−=∴

−=+=+=′⇒+=

⇒

∫∫∫∫

−−

−

11

Contour Integrals Ex. Evaluate , where C is the circle x = cos t, y = sin t, 0 ≤ t ≤ 2π. (sol)

∫C dzz1

( ) ( )( )( )

( ) idtidtieedzz

eztzfietzetittz

itit

C

it

itit

πππ

211

sincos

2

0

2

0===∴

===′⇒=+=

⇒

∫∫∫ −

−

12

Contour Integrals Properties of contour integrals Suppose f & g are continuous in a domain D

& C is a smooth curve lying entirely in D. ( ) ( ) ( )

( ) ( ) ( )[ ] ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )∫∫∫∫∫

∫∫∫∫∫

−=

+=+=

+=+

=

− CC

CCC

CCC

CC

dzzfdzzfiv

CCCdzzfdzzfdzzfiii

dzzgdzzfdzzgzfii

kdzzfkdzzkfi

,

constant. a ,

2121

13

Contour Integrals Ex. Evaluate , where C = C1 + C2.

C1 is defined by y = x = t, 0 ≤ t ≤ 1.

C2 is defined by x = 1, y = t, 1 ≤ t ≤ 2.

( )∫ +C

dziyx 22

( ) ( ) ( )∫∫∫ +++=+⇒21

222222

CCCdziyxdziyxdziyx

( ) ( )( )

( )321

1

1

0

22

1

0

2222

1

idtti

dtiittdziyxC

=+=

++=+

∫

∫∫

( ) ( ) idtidttidtitdziyxC

+−=+−=+=+ ∫∫∫∫ 371

2

1

2

1

22

1

222

2

( ) MLdzzfC

≤∫

Contour Integrals Thm. If f is continuous on a smooth curve C &

if |f (z)| ≤ M for all z on C, then , where L is the length of C.

(proof) From (5) of Ch.17, we can write

|∆zk|: length of the chord joining the points zk & zk−1.

( ) ( ) ∑∑∑===

∆≤∆≤∆n

kk

n

kkk

n

kkk zMzzfzzf

11

*

1

*

( ) MLzMzzfn

kk

n

kkk ≤∆≤∆⇒ ∑∑

== 11

*15

Contour Integrals Ex. Find an upper bound for , where

C is the circle |z| = 4. The length of the circle C is 8π. From the inequality of Ch.17,

∫ +C

z

dzze

1

311 =−≥+ zz( )

38

1

333sincos

3114

4

edzze

eeyiyeeze

ze

C

z

xxzzz

π≤

+∴

≤=+

==−

≤+

⇒

∫16

NT

( )( )

−=⋅

+=⋅

∫∫∫∫

5

4

CC

CC

dxvdyudsNf

dyvdxudsTf

Contour Integrals Circulation & net flux & denote the unit tangent vector & unit

normal vector to a positively oriented simple closed contour C. Interpret f (z) = u(x,y) + iv(x,y) as a vector, the line integrals

The line integral in (4) is called circulation around

C & measures the tendency of the flow to rotate C.

Contour Integrals The net flux across C is the difference between

the rate at which fluid enters & the rate at which fluid leaves the region bounded by C. It is given by the line integral in (5).

Note that

( ) ( ) ( )( ) ( )

( )( ) ( )( )( ) ( )

=

=∴

=+−=⋅+⋅

∫∫

∫∫∫∫

7Im

6Re

fluxNet

nCirculatio

C

C

CCCC

dzzf

dzzf

dzzfidydxivudsNfidsTf

18

Contour Integrals Ex. Given the flow f (z) = (1 + i)z, compute the

circulation around & the net flux across the circle C: |z| = 1. ( ) ( )( )( ) ( )

( ) ( )idti

dtieeidzzf

tetzzizf

itit

C

it

+=+=

−=⇒

≤≤=−=

⇒

∫∫∫ −

121

1

20 ,1

2

0

2

0

π

π

π

π

19

Cauchy-Goursat Theorem Focus on contour integrals where C is a

positive simple closed curve

Simply & multiply connected domains Simply connected if every simple

closed C lying entirely in D can be shrunk to a point without leaving D.

Ex. The entire complex plane. Multiply connected a domain that is not simply connected.

( ) 0=∫C dzzf

Cauchy-Goursat Theorem Cauchy’s Theorem Suppose that f is analytic in a simply connected

domain D & that f′ is continuous in D. For every simple closed contour C in D,

(proof) Since f′ is continuous throughout D, the real & imaginary parts of f (z) = u + iv & their 1st partial derivatives are continuous in D. From (2),

( ) ( ) ( ) ( ) ( )

0

,,,,

=

∂∂

−∂∂

+

∂∂

−∂∂

−=

++−=

∫∫∫∫

∫∫∫

CC DD

CCC

dAyv

xuidA

yu

xv

dyyxudxyxvidyyxvdxyxudzzf

21

Cauchy-Goursat Theorem Cauchy-Goursat Theorem Suppose f is analytic in a simply connected

domain D. Then for every simple closed contour C in D,

In other words, if f is analytic at all points within & on a simple closed contour C, then

Ex. Evaluate .

Since ez is entire & C is a simple closed contour, it follows from the Cauchy- Goursat theorem that

( ) ( )80 =∫C dzzf

∫Czdze

( ) 0=∫C dzzf

0=∫Czdze 22

Cauchy-Goursat Theorem Ex. Evaluate , where C:

f (z) = 1/z2 is analytic everywhere except at z = 0. But z = 0 is not interior to or on C.

Ex. Given the flow , compute the circulation around & net flux across C, where C is the square with vertices z = ±1 & z = ±i.

Circulation & net flux are both zero.

∫C zdz

2 ( ) ( ) 14

522

2 =−

+−yx

02 =∴∫C zdz

( ) zzf cos=

( ) 0 cos ==⇒ ∫∫ CCdzzdzzf

Cauchy-Goursat Theorem For multiply connected domains If f is analytic in a multiply connected domain D,

then we cannot conclude that for every simple closed C in D.

Suppose D is doubly connected and C & C1 are simple closed contours such that C1 surrounds the hole in D & is interior to C.

Also, suppose f is analytic on each contour & at each point interior to C but exterior to C1.

( ) 0=∫C dzzf

Cauchy-Goursat Theorem As we introduce the cut AB, the region bounded by

the curves is simply connected.

(9) is called the principle of deformation of contours. Thus, one can evaluate an integral over a complicated simple closed contour by replacing that contour with one that is more convenient.

( ) ( ) ( ) ( )

( ) ( ) ( )9

0

1

1

∫∫∫∫∫∫

=⇒

=+++⇒−

CC

BACABC

dzzfdzzf

dzzfdzzfdzzfdzzf

25

Cauchy-Goursat Theorem Ex. Evaluate , where C is the outer contour shown in the figure. From (9), we choose C1. Taking the radius of the circle to be r = 1, C1 lies within C.

Let z − i = eit, 0 ≤ t ≤ 2π

1:1 =−⇒ izC

∫ −C izdz

idteie

izdz

izdz

it

it

CCπ

π2

2

01

==−

=−

∴ ∫∫∫

itiez =′⇒

( )( )10

1 ,01 ,2

0

≠=

=−∫ n

nizz

dzC n

π

Cauchy-Goursat Theorem If z0 is any constant complex number interior

to any simple closed contour C, then For n is zero or negative, the integral is zero. For n is positive integer different from one,

( ) ( )( )

( ) 01

2

0

112

000

1

=−

==−

=−

∴−

−∫∫∫π

π

nieir

erdtire

zzdz

zzdz itn

nnitn

it

C nC n

itit ireztrezzC =′≤≤=−⇒ and , π20: 01

27

Cauchy-Goursat Theorem Ex. Evaluate with C: |z − 2| = 2.

∫ −++

Cdz

zzz

3275

2

( )( )

iizdz

zdzdz

zzz

zzzzz

zzz

CCC

ππ 60223 3

21

332

753

21

331

7532

75

2

2

=×+×=+

+−

=−+

+⇒

++

−=

+−+

=−+

+

∫∫∫

28

( ) ( ) ( )∫∫∫ +=⇒21 CCC

dzzfdzzfdzzf

Cauchy-Goursat Theorem If C, C1, C2 are simple closed contours & if f is

analytic on them & at points interior to C but exterior to both C1 & C2, then by introducing two cuts,

Cauchy-Goursat Theorem CG Thm for Multiply Connected Domains Suppose C, C1, …, Cn are positive simple closed

contours such that C1, C2, …, Cn are interior to C but their interior regions have no points in common. If f is analytic on each contour & at each point interior to C but exterior to all Ck, k = 1 … n,

( ) ( ) ( )111

∑∫∫=

=⇒n

kCC k

dzzfdzzf

30

Cauchy-Goursat Theorem Ex. Evaluate , where C: |z| = 3.

Surround z = ±i by circle contours C1 & C2 that lie entirely within C.

∫ +C zdz

12

∫∫

+−

−=

+∴

+−

−=

+⇒

CCdz

izizizdz

iziziz11

21

1

1121

11

2

2

21: and

21: 21 =+=−⇒ izCizC

31

02210

210

212

21

21

21

21

21

112111

21

1

2211

212

=⋅−⋅+⋅−⋅=

+−

−+

+−

−=

+−

−+

+−

−=

+⇒

∫∫∫∫

∫∫∫

iiii

ii

izdz

iizdz

iizdz

iizdz

i

dzizizi

dziziziz

dz

CCCC

CCC

ππ

Cauchy-Goursat Theorem

32

Cauchy-Goursat Theorem It can be shown that the Cauchy-Goursat

theorem is valid for any closed C in a simply connected domain D. As shown in the figure, if f is analytic in D, then

( ) 0=∫C dzzf

33

Independence of Path Def. Let z0 & z1 be points in D. A contour integral is said to be independent of the path if its value is the same for all contours C in D with an initial point z0 & a terminal point z1. Suppose C & C1 are contours in a simply

connected domain D, both from z0 to z1. Note that C & −C1 form a closed contour. Thus, if f is analytic in D,

( )∫C dzzf

( ) ( ) ( ) ( ) ( )12 0 11

∫∫∫∫ =∴=+⇒− CCCC

dzzfdzzfdzzfdzzf

Independence of Path Analyticity implies path independence

Thm. If f is an analytic function in a simply connected domain D, then is independent of C. Ex. Evaluate where C is the contour from z = −1 to z = −1 + i shown in the figure. Since f (z) = 2z is entire, we can replace C by any C1, say

( )∫C dzzf

∫C dzz 2

10 , ,1:1 ≤≤=−=⇒ ttyxC

( ) 12 12 2 21

01

−−=+−==∴ ∫∫∫ iidtitdzzdzzCC

35

Independence of Path A contour integral that is independent of C from z0 to z1 is written .

Def. Suppose f is continuous in D. If there exists a function F such that F′ (z) = f (z) for z in D, then F is called an antiderivative of f. The most general antiderivative of f (z) is written

Since F has a derivative at each point in D, it is necessarily analytic & hence continuous in D.

( )∫C dzzf ( )∫

1

0

z

zdzzf

( ) ( ) CzFdzzf +=∫

36

( ) ( ) ( ) ( )13 01 zFzFdzzfC

−=∫

Independence of Path Fundamental Theorem of Contour Integrals Thm. Suppose f is continuous in D & F is an

antiderivative of f in D. Then for any C in D from z0 to z1,

(proof) As C is a smooth curve defined by z = z(t), a ≤ t ≤ b. Use (3) & the fact that F′ (z) = f (z) for z in D,

( ) ( )( ) ( ) ( )( ) ( ) ( )( )

( )( ) ( )( ) ( )( ) ( ) ( )01 zFzFazFbzFtzF

dttzFdtddttztzFdttztzfdzzf

b

a

b

a

b

a

b

aC

−=−==

=′′=′= ∫∫∫∫37

Independence of Path Ex. Use (13) to solve the previous example. Ex. Evaluate , where C is any contour from z = 0 to z = 2 + i.

( ) ( ) ii

zdzzdzzii

C

2111

2222

1

1

21

1

−−=−−+−=

==⇒+−

−

+−

−∫∫

∫C dzz cos

( )i

i

zdzzdzz ii

C

4891.04031.1 0sin2sin

sin cos cos 2

0

2

0

−=−+=

== ++

∫∫

38

( ) ( )14 0 =∫C dzzf

Independence of Path

( )∫C dzzf

( )∫C dzzf

If C is closed, then z0 = z1 & consequently

Since the value of depends on only z0 & z1, it is the same for any C in D connecting these points. In other words, If f is continuous & has an antiderivative F in D, then is independent of C.

If f is continuous & is independent of C in D, then f has an antiderivative everywhere in D.

( )∫C dzzf

Independence of Path (proof) Assume that f is continuous, is independent of C in D, & F is a function defined by where s is a complex variable, z0 is a fixed point in D, & z is any point in D.

Choose ∆z so that z + ∆z is in D and z & z + ∆z can be joined by a straight segment in D.

( )∫C dzzf

( ) ( )∫≡z

zdssfzF

0

( ) ( ) ( ) ( )

( )∫

∫∫∆+

∆+

=

−=−∆+

zz

z

z

z

zz

z

dssf

dssfdssfzFzzF

00

40

Independence of Path (proof) As z fixed, we can write

Now f is continuous at the point z. It means that for any ε > 0 there exists a δ > 0 so that |f (s) − f (z)| < ε whenever |s − z| < δ.

( ) ( ) ( )

( ) ( )∫

∫∫∆+

∆+∆+

∆=⇒

==∆

zz

z

zz

z

zz

z

dszfz

zf

dszfdszfzzf

1

( ) ( ) ( ) ( ) ( )[ ]∫∆+

−∆

=−∆

−∆+∴

zz

zdszfsf

zzf

zzFzzF 1

41

Independence of Path (proof) Consequently, if we choose ∆z so that |∆z| < δ,

( ) ( ) ( ) ( ) ( )[ ]

( ) ( )[ ] εε =∆∆

≤−∆

=

−∆

=−∆

−∆+

∫

∫∆+

∆+

zz

dszfsfz

dszfsfz

zfz

zFzzF

zz

z

zz

z

11

1

( ) ( ) ( ) ( ) ( )zfzFzfz

zFzzFz

=′=∆

−∆+∴

→∆or lim

0

42

Independence of Path It is known that If f is analytic in a simply connected domain D, f is necessarily continuous throughout D. Its contour integral is independent of the path.

Combine with the result just obtained, Thm. If f is analytic in a simply connected

domain D, then f has an antiderivative F everywhere in D or there exists a function F so that F′ (z) = f (z) for all z in D.

43

Independence of Path Recall from Ch.17 that 1/z is the derivative of

Ln z. It means that under some circumstances Ln z is an antiderivative of 1/z. Suppose D is the entire complex plane without the

origin. 1/z is analytic in D. If C is any simple closed contour containing the origin,

In this case, Ln z is NOT an antiderivative of 1/z in D, since Ln z is not analytic in D. (Ln z fails to be analytic on the nonpositive real axis!)

idzzC

π21=∫

44

Independence of Path Ex. Evaluate with C shown in the figure. D is the simply connected domain defined by

Here, Ln z is an antiderivative of 1/z, since both Ln z & 1/z are analytic in D.

( ) ( ) 0Im ,0Re >=>= zyzx

∫C dzz1

( )

i

i

izdzz

ee

ii

5708.14055.0

3log2

2log

3Ln2LnLn 1 2

3

2

3

+−=

−+=

−==⇒ ∫π

45

Independence of Path If f & g are analytic in a simply connected

domain D containing C from z0 to z1, then the integration by parts formula is valid in D:

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )∫∫

∫∫∫

∫

′−=′∴

′+′=

=

1

0

1

0

1

0

1

0

1

0

1

0

z

z

z

z

z

z

z

z

z

zC

z

zC

dzzgzfzgzfdzzgzf

dzzgzfdzzgzfdzfgdzd

zgzfdzfgdzd

46

( ) ( ) ( )1521

00 ∫ −=

Cdz

zzzf

izf

π

Cauchy’s Integral Formulas Cauchy’s integral formula Thm. Let f be analytic in a simply connected

domain D and C be a simple closed contour lying entirely within D. If z0 is any point within C, then (proof) Let C1 be a circle centered at z0 with radius small enough that it is interior to C.

47

Cauchy’s Integral Formulas By the principle of deformation of contours,

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )∫

∫∫

∫

∫∫

−−

+⋅=

−−

+−

⋅=

−+−

=

−=

−⇒

1

11

1

1

0

00

0

0

00

0

00

00

2

C

CC

C

CC

dzzz

zfzfizf

dzzz

zfzfzz

dzzf

dzzz

zfzfzf

dzzzzfdz

zzzf

π48

Cauchy’s Integral Formulas Since f is continuous at z0, for any ε > 0, there exists a δ > 0 such that |f (z) − f (z0)| < ε whenever |z − z0| < δ. Thus, if we choose C1 to be |z − z0| = δ/2 < δ, then by the ML-inequality,

In other words, the absolute value of the integral can be made arbitrarily small by taking the radius of C1 to be sufficiently small. This can happen only if the integral is zero. (15) can thus be obtained.

( ) ( ) πεδπδε 2

22

21 0

0 =

⋅≤

−−

⇒ ∫C dzzz

zfzf

49

( ) ( ) ( )1521

00 ∫ −=⇒

Cdz

zzzf

izf

π

Cauchy’s Integral Formulas Since most problems do not have simply

connected domains, a more practical restatement of Cauchy’s integral theorem: If f is analytic within & on a simple closed contour C & z0 is interior to C,

Ex. Evaluate with C: |z| = 2. ∫ ++−

Cdz

izzz 442

( )

−=+−=

izzzzf

0

2 44 ( ) ( ) ( )iifidzzzzf

C3422

0

+−=−⋅=−

∴∫ ππ

Cauchy’s Integral Formulas Ex. Evaluate with C: |z − 2i| = 4.

( )

( ) iiiiifi

dziziz

z

dzz

z

iziz

zzfiziz

z

zz

CC

πππ =⋅=⋅=

−+=

+∴

=+

=⇒−+=

+

∫∫

63232

33

9

3 and 3

33

9

2

02

∫ +Cdz

zz

92

51

Cauchy’s Integral Formulas Ex. , where k = a + ib & z1 are complex numbers, gives rise to a flow in the domain z ≠ z1. If C is a simple closed contour containing z = z1 in its interior, then

Thus, the circulation around C is 2πb, & the net flux across C is 2πa.

If z1 were in the exterior of C, both the circulation & the net flux would be zero by Cauchy’s theorem.

( ) ( )1zzkzf −=

( ) ( )ibaidzzzibadzzf

CC−⋅=

−−

= ∫∫ π21

52

Cauchy’s Integral Formulas When k is real, the circulation around C is zero but

the net flux across C is 2πk. The complex number z1 is called a source for the flow when k > 0 & a sink when k < 0.

53

( )( ) ( )( )

( )162

!1

00 ∫ +−=

C nn dz

zzzf

inzfπ

Cauchy’s Integral Formulas Cauchy’s integral formula for derivatives Thm. Let f be analytic in a simply connected

domain D and C be a simple closed contour lying entirely within D. If z0 is any point within C, then

(proof) ( ) ( ) ( )

( )( )

( )

−

−∆+−∆

=

∆−∆+

=′

∫∫→∆

→∆

CCz

z

dzzzzfdz

zzzzf

zi

zzfzzfzf

000

00

00

21lim

lim

π

Cauchy’s Integral Formulas

Since f is continuous on C, there exists a real number M so that |f (z)| ≤ M for z on C. Besides, let L be the length of C and δ be the shortest distance between points on C & z0. Thus, for all z on C,

If we choose |∆z| ≤ δ/2, then

( ) ( )( )( )

−∆−−

=′⇒ ∫→∆ Czdz

zzzzzzf

izf

0000 lim

21π

220

011or δ

δ ≤−

≥−⇒zz

zz

δδδ 21 and 2 0

00 ≤∆−−

≥∆−≥∆−−≥∆−−zzz

zzzzzzz

Cauchy’s Integral Formulas

Since the last expression goes to zero as ∆z → 0,

( )( )( )

( )( )

( )( ) ( ) 3

02

0

2000

2δ

zMLdz

zzzzzzfz

dzzzzfdz

zzzzzzf

C

CC

∆≤

∆−−−⋅∆

=

−−

−∆−−

∫

∫∫

( ) ( )( )∫ −

=′∴C

dzzzzf

izf 2

00 2

1π

56

Cauchy’s Integral Formulas If f (z) = u(x,y) + iv(x,y) is analytic at a point,

then its derivatives of all orders exist at that point & are continuous. Consequently, from

u & v have continuous partial derivatives of all orders at a point of analyticity.

( )

( )

xyui

xyv

xvi

xuzf

yui

yv

xvi

xuzf

∂∂∂

−∂∂

∂=

∂∂

+∂∂

=′′

∂∂

−∂∂

=∂∂

+∂∂

=′

22

2

2

2

2

57

Cauchy’s Integral Formulas Ex. Evaluate with C: |z| = 1.

The integrand is not analytic at z = 0 & z = −4, but only z = 0 lies within C.

By (16) we have

∫ ++

Cdz

zzz

34 41

( ) 2 ,0 ,41 4

1

41

0334 ==++

=⇒++

=++

⇒ nzzzzf

zzz

zzz

( )( )

iz

ifidzzz

z

zC 32

3460

!22

41

0334

πππ−=

+−

⋅=′′=++

=∫

58

Cauchy’s Integral Formulas Ex. Evaluate with C shown in the figure. Think of C as the union of two simple closed

contours −C1 & C2.

( )∫ −+

Cdz

izzz

2

3 3

( ) ( ) ( )

( )( )

21

2

3

2

3

2

3

2

3

2

3

33

333

21

21

II

dziz

zz

dzziz

z

dzizz

zdzizz

zdzizz

z

CC

CCC

+−=−

+

+−+

−=

−+

+−+

=−+

∫∫

∫∫∫ −

59

Cauchy’s Integral Formulas Use (15) & (16) for I1 & I2, respectively.

( )( )

( )( )

( )iIIdz

izzz

iz

ziifidziz

zz

I

iiz

zidzziz

z

I

C

izC

zC

ππ

ππππ

ππ

1243

64322!1

23

632

3

212

3

2

3

2

3

2

02

32

3

1

2

1

+−=+−=−+

∴

+−=−

⋅=′=−

+

=

−=−+

⋅=−+

=

∫

∫

∫

=

=

60

Cauchy’s Integral Formulas Liouville’s Theorem The only bounded entire functions are constants.

Suppose f is entire & bounded, that is, |f (z)| ≤ M for all z. Then for any point z0, |f′ (z0)| ≤ M/r.

By taking r arbitrarily large, we can make |f′ (z0)| as small as we wish. f′ (z0) = 0 for all points z0 in the complex plane.

Hence f must be a constant.

( )( ) ( )( ) nnC n

n

rMnr

rMndz

zzzfnzf !2

2!

2!

110

0 =≤−

= ++∫ πππ

61

Cauchy’s Integral Formulas Fundamental theorem of algebra If P(z) is a nonconstant polynomial, then the equation P(z) = 0 has at least one root. Suppose that P(z) ≠ 0 for all z. This implies that the

reciprocal of P, f (z) = 1/P(z), is an entire function. Since |f (z)| → 0 as |z| → ∞, f must be bounded for

all finite z. It follows from Liouville’s thm that f is a constant

& therefore P is a constant. Contradiction! We conclude that there must exist at least one

number z for which P(z) = 0 . 62

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