Integración variable compleja. Integral de Cauchy

Integración compleja

1. 4. Complex integration: Cauchy integral theorem and Cauchy
integral formulas
Definite integral of a complex-valued function of a real variable
Consider a complex valued functionf(t) of a real variablet:
f(t) =u(t)+iv(t),
which is assumed to be a piecewise continuous function define d in
the closed intervala≤t≤b. The integral off(t) fromt=ato
t=b, is defined as
Z
b
a
f(t)dt=
Z
b
a
u(t)dt+i
Z
b
a
v(t)dt.
1

2. Properties of a complex integral with real variable of integration
1.
Re
Z
b
a
f(t)dt=
Z
b
a
Ref(t)dt=
Z
b
a
u(t)dt.
2.
Im
Z
b
a
f(t)dt=
Z
b
a
Imf(t)dt=
Z
b
a
v(t)dt.
3.
Z
b
a
[γ1f1(t)+γ2f2(t)]dt=γ1
Z
b
a
f1(t)dt+γ2
Z
b
a
f2(t)dt,
whereγ1andγ2are any complex constants.
4.






Z
b
a
f(t)dt






≤
Z
b
a
|f(t)|dt.
2

3. To prove (4), we consider






Z
b
a
f(t)dt






=e
−iφ
Z
b
a
f(t)dt=
Z
b
a
e
−iφ
f(t)dt,
whereφ= Arg

Z
b
a
f(t)dt
!
. Since






Z
b
a
f(t)dt






is real, we deduce that






Z
b
a
f(t)dt






= Re
Z
b
a
e
−iφ
f(t)dt=
Z
b
a
Re [e
−iφ
f(t)]dt
≤
Z
b
a
|e
−iφ
f(t)|dt=
Z
b
a
|f(t)|dt.
3

4. Example
Supposeαis real, show that
|e
2απi
−1| ≤2π|α|.
Solution
Letf(t) =e
iαt
,αandtare real. We obtain






Z
2π
0
e
iαt
dt






≤
Z
2π
0
|e
iαt
|dt= 2π.
The left-hand side of the above inequality is equal to






Z
2π
0
e
iαt
dt






=






e
iαt
iα






2π
0






=
|e
2απi
−1|
|α|
.
Combining the results, we obtain
|e
2απi
−1| ≤2π|α|, αis real.
4

5. Definition of a contour integral
Consider a curveCwhich is a set of pointsz= (x,y) in the complex
plane defined by
x=x(t), y=y(t), a≤t≤b,
wherex(t) andy(t) are continuous functions of the real parameter
t. One may write
z(t) =x(t)+iy(t), a≤t≤b.
•The curve is said to besmoothifz(t) has continuous derivative
z
′
(t)6= 0 for all points along the curve.
•Acontouris defined as a curve consisting of a finite number
of smooth curves joined end to end. A contour is said to be a
simple closed contourif the initial and final values ofz(t) are
the same and the contour does not cross itself.
5

6. •Letf(z) be any complex function defined in a domain Din the
complex plane and letCbe any contour contained inDwith
initial pointz0and terminal pointz.
•We divide the contourCintonsubarcs by discrete pointsz0, z1, z2,
...,zn−1,zn=zarranged consecutively along the direction of in-
creasingt.
•Letζ
kbe an arbitrary point in the subarcz
kz
k+1and form the
sum
n−1
X
k=0
f(ζ
k)(z
k+1−z
k).
6

7. Subdivision of the contour intonsubarcs by discrete pointsz0,z1,∆∆∆,
zn−1,zn=z.
7

8. We write△z
k=z
k+1−z
k. Letλ= max
k
|△z
k|and take the limit
lim
λ→0
n→∞
n−1
X
k=0
f(ζ
k)△z
k.
The above limit is defined to be thecontour integraloff(z) along
the contourC.
If the above limit exists, then the functionf(z) is said to beinte-
grable along the contourC.
If we write
dz(t)
dt
=
dx(t)
dt
+i
dy(t)
dt
, a≤t≤b,
then
Z
C
f(z)dz=
Z
b
a
f(z(t))
dz(t)
dt
dt.
8

9. Writingf(z) =u(x,y)+iv(x,y) anddz=dx+idy, we have
Z
C
f(z)dz=
Z
C
u dx−v dy+i
Z
C
u dy+v dx
=
Z
b
a
"
u(x(t),y(t))
dx(t)
dt
−v(x(t),y(t))
dy(t)
dt
#
dt
+i
Z
b
a
"
u(x(t),y(t))
dy(t)
dt
+v(x(t),y(t))
dx(t)
dt
#
dt.
The usual properties of real line integrals are carried overto their
complex counterparts. Some of these properties are:
(i)
Z
C
f(z)dzis independent of the parameterization ofC;
(ii)
Z
−C
f(z)dz=−
Z
C
f(z)dz, where−Cis the opposite curve ofC;
(iii) The integrals off(z) along a string of contours is equal to the
sum of the integrals off(z) along each of these contours.
9

10. Example
Evaluate the integral
I
C
1
z−z0
dz,
whereCis a circle centered atz0and of any radius. The path is
traced out once in the anticlockwise direction.
Solution
The circle can be parameterized by
z(t) =z0+re
it
,0≤t≤2π,
whereris any positive real number. The contour integral becomes
I
C
1
z−z0
dz=
Z
2π
0
1
z(t)−z0
dz(t)
dt
dt=
Z
2π
0
ire
it
re
it
dt= 2πi.
The value of the integral is independent of the radiusr.
10

11. Example
Evaluate the integral
(i)
Z
C
|z|
2
dzand (ii)
Z
C
1
z
2
dz,
where the contourCis
(a) the line segment with initial point−1 and final pointi;
(b) the arc of the unit circle Imz≥0 with initial point−1 and final
pointi.
Do the two results agree?
11

12. Solution
(i) Consider
Z
C
|z|
2
dz,
(a) Parameterize the line segment by
z=−1+(1+ i)t, 0≤t≤1,
so that
|z|
2
= (−1+t)
2
+t
2
and dz= (1+i)dt.
The value of the integral becomes
Z
C
|z|
2
dz=
Z
1
0
(2t
2
−2t+1)(1+i)dt=
2
3
(1+i).
12

13. (b) Along the unit circle,|z|= 1 andz=e
iθ
,dz=ie
iθ
dθ. The initial
point and the final point of the path correspond to θ=πand
θ=
π
2
, respectively. The contour integral can be evaluated as
Z
C
|z|
2
dz=
Zπ
2
π
ie
iθ
dθ=e
iθ






π
2π
= 1+i.
The results in (a) and (b) do not agree. Hence, the value of this
contour integral does depend on the path of integration.
13

14. (ii) Consider
Z
C
1
z
2
dz.
(a) line segment from−1 toi
Z
C
1
z
2
dz=
Z
1
0
1+i[−1+(1+ i)t]
2
dt=−
1
−1+(1+ i)t





1
0
=−1−
1
i
=−1+i.
(b) subarc from−1 toi
Z
C
1
z
2
dz=
Zπ
2
π
1e
2iθ
ie
iθ
dθ=−e
−iθ
λπ
2π
=−1+i.
14

15. Estimation of the absolute value of a complex integral
The upper bound for the absolute value of a complex integral c an
be related to the length of the contourCand the absolute value of
f(z) alongC. In fact,






Z
C
f(z)dz






≤ML,
whereMis the upper bound of|f(z)|alongCandLis the arc length
of the contourC.
15

16. We consider






Z
C
f(z)dz






=






Z
b
a
f(z(t))
dz(t)
dt
dt






≤
Z
b
a
|f(z(t))|






dz(t)
dt






dt
≤
Z
b
a
M






dz(t)
dt






dt
=M
Z
b
a
v
u
u
t

dx(t)
dt
!
2
+

dy(t)
dt
!
2
dt=ML.
16

17. Example
Show that





Z
C
1
z
2
dz






≤2, whereCis the line segment joining−1+iand 1+i.
Solution
Along the contourC, we havez=x+i,−1≤x≤1, so that 1≤
|z| ≤
√
2. Correspondingly,
1
2
≤
1
|z|
2
≤1. Here,M= max
z∈C
1
|z|
2
= 1
and the arc lengthL= 2. We have






Z
C
1
z
2
dz






≤ML= 2.
17

18. Example
Estimate an upper bound of the modulus of the integral
I=
Z
C
Logz
z−4i
dz
whereCis the circle|z|= 3.
Now,






Logz
z−4i






≤





ln|z|





+|Argz|
||z|−|4i||
so that
max
z∈C






Logz
z−4i






≤
ln3+π
|3−4|
= ln3+π;L= (2π)(3) = 6π.
Hence,






Z
C
Logz
z−4i
dz






≤6π(π+ln3).
18

19. Example
Find an upper bound for





Z
Γ
e
z
/(z
2
+1)dz





, where Γ is the circle
|z|= 2 traversed once in the counterclockwise direction.
Solution
The path of integration has lengthL= 4π. Next we seek an upper
boundMfor the functione
z
/(z
2
+ 1) when|z|= 2. Writingz=
x+iy, we have
|e
z
|=|e
x+iy
|=e
x
≤e
2
,for|z|=
q
x
2
+y
2
= 2,
and by the triangle inequality
|z
2
+1| ≥ |z|
2
−1 = 4−1 = 3 for |z|= 2.
Hence,|e
z
/(z
2
+1)| ≤e
2
/3 for|z|= 2, and so





Z
Γ
e
z
z
2
+1
dz





≤
e
2
3
∆4π.
19

20. Path independence
Under what conditions that
Z
C
1
f(z)dz=
Z
C
2
f(z)dz,
whereC1andC2are two contours in a domain Dwith the same
initial and final points andf(z) is piecewise continuous insideD.
The property of path independence is valid forf(z) =
1
z
2
but it fails
whenf(z) =|z|
2
. The above query is equivalent to the question:
When does
I
C
f(z)dz= 0
hold, whereCis any closed contour lying completely insideD? The
equivalence is revealed if we treatCasC1∪−C2.
We observe thatf(z) =
1
z
2
is analytic everywhere except atz= 0
butf(z) =|z|
2
is nowhere analytic.
20

21. Cauchy integral theorem
Letf(z) =u(x,y)+iv(x,y)be analytic on and inside a simple closed
contourCand letf
′
(z)be also continuous on and inside C, then
I
C
f(z)dz= 0.
Proof
The proof of the Cauchy integral theorem requires the Green theo-
rem for a positively oriented closed contourC: If the two real func-
tionsP(x,y) andQ(x,y) have continuous first order partial deriva-
tives on and insideC, then
I
C
P dx+Q dy=
ZZ
D
(Qx−Py)dxdy,
whereDis the simply connected domain bounded by C.
21

22. Suppose we writef(z) =u(x,y)+iv(x,y),z=x+iy; we have
I
C
f(z)dz=
I
C
u dx−v dy+i
I
C
v dx+u dy.
One can infer from the continuity off
′
(z) thatu(x,y) andv(x,y)
have continuous derivatives on and insideC. Using the Green the-
orem, the two real line integrals can be transformed into dou ble
integrals.
I
C
f(z)dz=
ZZ
D
(−vx−uy)dxdy+i
ZZ
D
(ux−vy)dxdy.
Both integrands in the double integrals are equal to zero dueto the
Cauchy-Riemann relations, hence the theorem.
In 1903, Goursat was able to obtain the same resultwithout assum-
ing the continuity off
′
(z).
22

23. Goursat Theorem
If a functionf(z) is analytic throughout a simply connected domain
D, then for any simple closed contourClying completely insideD,
we have
I
C
f(z)dz= 0.
Corollary 1
The integral of a functionf(z) which is analytic throughout a simply
connected domain Ddepends on the end points and not on the
particular contour taken. Supposeαandβare insideD,C1andC2
are any contours insideDjoiningαtoβ, then
Z
C
1
f(z)dz=
Z
C
2
f(z)dz.
23

24. Example
IfCis the curvey=x
3
−3x
2
+4x−1 joining points (1,1) and (2,3),
find the value of
Z
C
(12z
2
−4iz)dz.
Method 1. The integral is independent of the path joining (1,1)
and (2,3). Hence any path can be chosen. In particular, let us
choose the straight line paths from (1,1) to (2,1) and then from
(2,1) to (2,3).
Case 1Along the path from (1,1) to (2,1),y= 1,dy= 0 so that
z=x+iy=x+i,dz=dx. Then the integral equals
Z
2
1
{12(x+i)
2
−4i(x+i)}dx={4(x+i)
3
−2i(x+i)
2
}





2
1
= 20+30i.
24

25. Case 2Along the path from (2,1) to (2,3),x= 2,dx= 0 so that
z=x+iy= 2+iy,dz=idy. Then the integral equalsZ
3
1
{12(2+iy)
2
−4i(2+iy)}i dy={4(2+iy)
3
−2i(2+iy)
2
}





3
1
=−176+8i.
Then adding, the required value = (20 + 30 i) + (−176 + 8i) =
−156+38i.
Method 2. The given integral equals
Z
2+3i
1+i
(12z
2
−4iz)dz= (4z
3
−2iz
2
)





2+3i
1+i
=−156+38i.
It is clear that Method 2 is easier.
25

26. Corollary 2
Letf(z) be analytic throughout a simply connected domainD. Con-
sider a fixed pointz0∈ D; by virtue of Corollary 1,
F(z) =
Z
z
z0
f(ζ)dζ,for anyz∈ D,
is a well-defined function inD. Considering
F(z+∆z)−F(z)
∆z
−f(z) =
1
∆z
Z
z+∆z
z
[f(ζ)−f(z)]dζ.
By the Cauchy Theorem, the last integral is independent of the path
joiningzandz+∆zso long as the path is completely insideD. We
choose the path as the straight line segment joiningzandz+ ∆z
and choose|∆z|small enough so that it is completely insideD.
26

27. By continuity off(z), we have for all pointsuon this straight line
path
|f(u)−f(z)|< ǫwhenever|u−z|< δ.
Note that|∆z|< δis observed implicitly.
27

28. We have





Z
z+∆z
z
[f(u)−f(z)]du





< ǫ|∆z|
so that





F(z+∆z)−F(z)
∆z
−f(z)





=
1
|∆z|





Z
z+∆z
z
[f(u)−f(z)]du





< ǫ
for|∆z|< δ. This amounts to say
lim
∆z→0
F(z+∆z)−F(z)
∆z
=f(z),
that is,F
′
(z) =f(z) for allzinD. Hence,F(z) is analytic inD
sinceF
′
(z) exists at all points inD(which is an open set).
28

29. •This corollary may be considered as a complex counterpart of
the fundamental theorem of real calculus.
•If we integratef(z) along any contour joiningαandβinsideD,
then the value of the integral is given by
Z
β
α
f(z)dz=
Z
β
z0
f(ζ)dζ−
Z
α
z0
f(ζ)dζ
=F(β)−F(α), αandβ∈ D.
29

30. Corollary 3
LetC, C1, C2,...,Cnbe positively oriented closed contours, where
C1, C2,...,Cnare all insideC. ForC1, C2,...,Cn, each of these
contours lies outside of the other contours. Let intC
idenote the
collection of all points bounded insideCi. Letf(z) be analytic on
the setS:C∪intC\intC1\intC2n∆∆∆nintCn(see the shaded
area in Figure), then
I
C
f(z)dz=
n
X
k=1
I
C
k
f(z)dz.
30

31. The proof for the case whenn= 2 is presented below.
31

32. Proof
•The constructed boundary curve is composed of C∪−C1∪−C2
and the cut lines, each cut line travels twice in opposite direc-
tions.
•To explain the negative signs in front ofC1andC2, we note that
the interior contours traverse in the clockwise sense as parts of
the positively oriented boundary curve.
•With the introduction of these cuts, the shaded region bounded
within this constructed boundary curve becomes a simply con -
nected set.
We have
I
C
f(z)dz+
Z
−C
1
f(z)dz+
Z
−C
2
f(z)dz= 0,
so that
I
C
f(z)dz=
Z
C
1
f(z)dz+
Z
C
2
f(z)dz.
32

33. Example
LetDbe the domain that contains the whole complex plane except
the origin and the negative real axis. Let Γ be an arbitrary contour
lying completely insideD, and Γ starts from 1 and ends at a point
α. Show that
Z
Γ
dz
z
= Logα.
Solution
Let Γ1be the line segment from 1 to|α|along the real axis, and
Γ2be a circular arc centered at the origin and of radius|α|which
extends from|α|toα. The union Γ 1∪Γ2∪ −Γ forms a closed
contour. Since the integrand
1
z
is analytic everywhere insideD, by
the Cauchy integral theorem, we have
Z
Γ
dz
z
=
Z
Γ
1
dz
z
+
Z
Γ
2
dz
z
.
33

34. The contour Γ starts fromz= 1 and ends atz=α. The arc Γ2is
part of the circle|z|=|α|.
34

35. Sinceαcannot lie on the negative real axis, so Argαcannot assume
the valueπ. If we writeα=|α|e
iArgα
(−π <Argα < π), then
Z
Γ
1
dz
z
=
Z
|α|
1
dt
t
= ln|α|
Z
Γ
2
dz
z
=
Z
Argα
0
ire
iθ
re
iθ
dθ=iArgα.
Combining the results,
Z
Γ
dz
z
= ln|α|+iArgα= Logα.
Note that the given domainDis the domain of definition of Logz,
the principal branch of the complex logarithm function.
35

36. Poisson integral
Consider the integration of the functione
−z
2
around the rectangular
contour Γ with vertices±a,±a+iband oriented positively. By letting
a→ ∞while keepingbfixed, show that
Z
∞
−∞
e
−x
2
e
±2ibx
dx=
Z
∞
−∞
e
−x
2
cos2bx dx=e
−b
2
√
π.
y
x
(
a
,
b
)
(
-
a
,
b
)
(
-
a
, 0)
(
a
, 0)
G
1
G
2
G
3
G
4
The configuration of the closed rectangular contour Γ.
36

37. Solution
Sincee
−z
2
is an entire function, we have
I
Γ
e
−z
2
dz= 0,
by virtue of the Cauchy integral theorem. The closed contour Γ
consists of four line segments: Γ = Γ1∪Γ2∪Γ3∪Γ4, where
Γ1={x:−a≤x≤a},
Γ2={a+iy: 0≤y≤b},
Γ3={x+ib:−a≤x≤a},
Γ4={−a+iy: 0≤y≤b},
and Γ is oriented in the anticlockwise direction.
37

38. The contour integral can be split into four contour integrals, namely,
I
Γ
e
−z
2
dz=
Z
Γ
1
e
−z
2
dz+
Z
Γ
2
e
−z
2
dz+
Z
Γ
3
e
−z
2
dz+
Z
Γ
4
e
−z
2
dz.
The four contour integrals can be expressed as real integral s as
follows:
Z
Γ
1
e
−z
2
dz=
Z
a
−a
e
−x
2
dx,
Z
Γ
2
e
−z
2
dz=
Z
b
0
e
−(a+iy)
2
i dy,
Z
Γ
3
e
−z
2
dz=
Z
−a
a
e
−(x+ib)
2
dx,
=−e
b
2
≤Z
a
−a
e
−x
2
cos2bx dx−i
Z
a
−a
e
−x
2
sin2bx dx
λ
,
Z
Γ
4
e
−z
2
dz=
Z
0
b
e
−(−a+iy)
2
i dy.
38

39. First, we consider the bound on the modulus of the second integral.






Z
Γ
2
e
−z
2
dz






≤
Z
b
0
|e
−(a
2
−y
2
+2iay)
i|dy
=e
−a
2
Z
b
0
e
y
2
dy
≤e
−a
2
Z
b
0
e
b
2
dy (since 0≤y≤b)
=
be
b
2
e
a
2
→0 asa→ ∞andbis fixed.
Therefore, the value of
Z
Γ
2
e
−z
2
dz→0 asa→ ∞.
By similar argument, the fourth integral can be shown to be ze ro
asa→ ∞.
39

40. lim
a→∞
I
Γ
e
−z
2
dz= lim
a→∞
θZ
a
−a
e
−x
2
dx−e
b
2
Z
a
−a
e
−x
2
cos2bx dx

+ilim
a→∞
θ
e
b
2
Z
a
−a
e
−x
2
sin2bx dx

= 0,
so that
Z
∞
−∞
e
−x
2
cos2bx dx−i
Z
∞
−∞
e
−x
2
sin2bx dx=e
−b
2
Z
∞
−∞
e
−x
2
dx=e
−b
2
√
π.
Either by equating the imaginary parts of the above equation or
observing thate
−x
2
sin2bxis odd, we deduce
Z
∞
−∞
e
−x
2
sin2bx dx= 0.
Hence, we obtain
Z
∞
−∞
e
−x
2
e
±2ibx
dx=
Z
∞
−∞
e
−x
2
cos2bx dx=e
−b
2
√
π.
40

41. Cauchy integral formula
Let the functionf(z)be analytic on and inside a positively oriented
simple closed contourCandzis any point insideC, then
f(z) =
1
2πi
I
C
f(ζ)
ζ−z
dζ.
Proof
We draw a circleCr, with radiusraround the pointz, small enough
to be completely insideC. Since
f(ζ)
ζ−z
is analytic in the region lying
betweenCrandC, we have
1
2πi
I
C
f(ζ)
ζ−z
dζ=
1
2πi
I
C
r
f(ζ)ζ−z
dζ
=
1
2πi
I
C
r
f(ζ)−f(z)
ζ−z
dζ+
f(z)
2πi
I
C
r
1
ζ−z
dζ.
The last integral is seen to be equal tof(z). To complete the proof,
it suffices to show that the first integral equals zero.
41

42. ´
x
y
z
C
C
r
The contourCis deformed to the circleCr, which encircles the
pointz.
42

43. Sincefis continuous atz, for eachǫ >0, there existsδ >0 such
that
|f(ζ)−f(z)|< ǫwhenever|ζ−z|< δ.
Now, suppose we choose r < δ(it is necessary to guarantee that
Crlies completely inside the contourC), the modulus of the first
integral is bounded by





1
2πi
I
C
r
f(ζ)−f(z)
ζ−z
dζ





≤
1
2π
I
C
r
|f(ζ)−f(z)|
|ζ−z|
|dζ|
=
1
2πr
I
C
r
|f(ζ)−f(z)| |dζ|
<
ǫ
2πr
I
C
r
|dζ|=
ǫ
2πr
2πr=ǫ.
Since the modulus of the above integral is less than any positive
numberǫ, however small, so the value of that integral is zero.
43

44. By the Cauchy integral formula, the value off(z) at any point inside
the closed contourCis determined by the values of the function
along the boundary contourC.
Example
Apply the Cauchy integral formula to the integral
I
|z|=1
e
kz
z
dz, kis a real constant,
to show that
Z
2π
0
e
kcosθ
sin(ksinθ)dθ= 0
Z
2π
0
e
kcosθ
cos(ksinθ)dθ= 2π.
44

45. Solution
By Cauchy’s integral formula:
I
|z|=1
e
kz
z
dz= (2πi)e
kz






z=0
= 2πi.
On the other hand,
2πi=
I
|z|=1
e
kz
z
dz=
Z
2π
0
e
k(cosθ+isinθ)
e
iθ
ie
iθ
dθ
=i
Z
2π
0
e
kcosθ
[cos(ksinθ)+isin(ksinθ)]dθ.
Equating the real and imaginary parts, we obtain
0 =
Z
2π
0
e
kcosθ
sin(ksinθ)dθ
2π=
Z
2π
0
e
kcosθ
cos(ksinθ)dθ.
45

46. Example
Evaluate
I
C
sinπz
2
+cosπz
2
(z−1)(z−2)
dz,
whereCis the circle:|z−i|= 3.
Solution
I
C
sinπz
2
+cosπz
2
(z−1)(z−2)
dz=
I
C
sinπz
2
+cosπz
2
z−2
dz−
I
C
sinπz
2
+cosπz
2
z−1
dz
By Cauchy’s integral formula, we have
I
C
sinπz
2
+cosπz
2
z−2
dz= 2πi{sinπ(2)
2
+cosπ(2)
2
}= 2πi
I
C
sinπz
2
+cosπz
2
z−1
dz= 2πi{sinπ(1)
2
+cosπ(1)
2
}=−2πi
sincez= 1 andz= 2 are insideCand sinπz
2
+ cosπz
2
is analytic
on and insideC. The integral has the value 2πi−(−2πi) = 4πi.
46

47. Remark
Alternately, by Corollary 3 of the Cauchy Integral Theorem,we have
I
C
sinπz
2
+cosπz
2
(z−1)(z−2)
dz=
I
C
1
(sinπz
2
+cosπz
2
)/(z−2)
z−1
dz
+
I
C
2
(sinπz
2
+cosπz
2
)/(z−1)
z−2
dz,
whereC1andC2are closed contours completely insideC,C1encir-
cles the pointz= 1 whileC2encircles the pointz= 2.
47

48. By the Cauchy Integral formula, choosingf(z) =
sinπz
2
+cosπz
2
z−2
,
we obtain
I
C
1
f(z)
z−1
dz= 2πif(1) = 2πi
sinπ+cosπ
−1
= 2πi.
In a similar manner
I
C
2
(sinπz
2
+cosπz
2
)/(z−1)
z−2
dz= 2πi
sinπz
2
+cosπz
2
z−1





z=2
= 2πi.
Hence, the integral is equal to 2πi+2πi= 4πi.
48

49. The Cauchy integral formula can be extended to the case where the
simple closed contourCcan be replaced by the oriented boundary
of a multiply connected domain.
SupposeC, C1, C2,...,Cnandf(z) are given the same conditions
as in Corollary 3, then for any pointz∈C∪intC\intC1\
intC2n∆∆∆nintCn, we have
f(z) =
1
2πi
I
C
f(ζ)
ζ−z
dζ−
n
X
k=1
1
2πi
I
C
k
f(ζ)ζ−z
dζ.
49

50. Derivatives of contour integrals
Suppose we differentiate both sides of the Cauchy integral formula
formally with respect toz(holdingζfixed), assuming that differen-
tiation under the integral sign is legitimate, we obtain
f
′
(z) =
1
2πi
d
dz
I
C
f(ζ)
ζ−z
dζ=
1
2πi
I
C
d
dz
f(ζ)ζ−z
dζ=
1
2πi
I
C
f(ζ)
(ζ−z)
2
dζ.
How to justify the legitimacy of direct differentiation of the Cauchy
integral formula? First, consider the expression
f(z+h)−f(z)
h
−
1
2πi
I
C
f(ζ)
(ζ−z)
2
dζ
=
1
h
(
1
2πi
I
C
"
f(ζ)
ζ−z−h
−
f(ζ)
ζ−z
−h
f(ζ)
(ζ−z)
2
#
dζ
)
=
h
2πi
I
C
f(ζ)
(ζ−z−h) (ζ−z)
2
dζ.
50

51. It suffices to show that the value of the last integral goes to zero
ash→0. To estimate the value of the last integral, we draw the
circleC
2d:|ζ−z|= 2dinside the domain bounded byCand choose
hsuch that 0<|h|< d.
For every pointζon the curveC, it is outside the circleC
2dso that
|ζ−z|> dand|ζ−z−h|> d.
LetMbe the upper bound of |f(z)|onCandLis the total arc
length ofC. Using the modulus inequality and together with the
above inequalities, we obtain






h
2πi
I
C
f(ζ)
(ζ−z−h) (ζ−z)
2
dζ






≤
|h|
2π
ML
d
3
.
51

52. In the limith→0, we observe that
lim
h→0






h
2πi
I
C
f(ζ)
(ζ−z−h)(ζ−z)
2
dζ






≤lim
h→0
|h|
2π
ML
d
3
= 0;
therefore,
f
′
(z) = lim
h→0
f(z+h)−f(z)
h
=
1
2πi
I
C
f(ζ)
(ζ−z)
2
dζ.
By induction, we can show the general result
f
(k)
(z) =
k!
2πi
I
C
f(ζ)
(ζ−z)
k+1
dζ, k = 1,2,3,∆∆∆,
for anyzinsideC. This result is called thegeneralized Cauchy
Integral Formula.
52

53. Theorem
If a functionf(z)is analytic at a point, then its derivatives of all
orders are also analytic at the same point.
Proof
Supposefis analytic at a pointz0, then there exists a neighborhood
ofz0:|z−z0|< ǫthroughout whichfis analytic. TakeC0to be
a positively oriented circle centered atz0and with radiusǫ/2 such
thatfis analytic inside and onC0. We then have
f
′′
(z) =
1
πi
I
C
0
f(ζ)
(ζ−z)
3
dζ
at each pointzinterior toC0. The existence off
′′
(z) throughout
the neighborhood:|z−z0|< ǫ/2 means thatf
′
is analytic atz0.
Repeating the argument to the analytic functionf
′
, we can conclude
thatf
′′
is analytic atz0.
53

54. Remarks
(i) The above theorem is limited to complex functions only. I n
fact, no similar statement can be made on real differentiable
functions. It is easy to find examples of real valued function
f(x) such thatf
′
(x) exists but not so forf
′′
(x) at certain points.
(ii) Suppose we express an analytic function inside a domain Das
f(z) =u(x,y) +iv(x,y),z=x+iy. Since its derivatives of
all orders are analytic functions, it then follows that the par-
tial derivatives ofu(x,y) andv(x,y) of all orders exist and are
continuous.
54

55. To see this, sincef
′′
(z) exists, we consider
f
′′
(z) =
∂
2
u
∂x
2
+i
∂
2
v
∂x
2
=
∂
2
v
∂y∂x
−i
∂
2
u
∂y∂x
≤
fromf
′
(z) =
∂u
∂x
+i
∂v
∂x
λ
or
f
′′
(z) =
∂
2
v
∂x∂y
−i
∂
2
u
∂x∂y
=−
∂
2
u
∂y
2
−i
∂
2
v
∂y
2
"
fromf
′
(z) =
∂v
∂y
−i
∂u
∂y
#
.
The continuity off
′′
implies that all second order partials ofuand
vare continuous at points wherefis analytic. Continuing with the
process, we obtain the result.
The mere assumption of the analyticity off(z) on and insideCis
sufficient to guarantee the existence of the derivatives off(z) of all
orders. Moreover, the derivatives are all continuous on andinside
C.
55

56. Example
Supposef(z) is defined by the integral
f(z) =
I
|ζ|=3
3ζ
2
+7ζ+1
ζ−z
dζ,
findf
′
(1+i).
Solution
Settingk= 1 in the generalized Cauchy integral formula,
f
′
(z) =
I
|ζ|=3
3ζ
2
+7ζ+1
(ζ−z)
2
dζ
=
I
|ζ|=3
3(ζ−z)
2
+(6z+7)(ζ−z)+3z
2
+7z+1
(ζ−z)
2
dζ
=
I
|ζ|=3
3dζ+(6z+7)
I
|ζ|=3
1
ζ−z
dζ
+ (3z
2
+7z+1)
I
|ζ|=3
1
(ζ−z)
2
dζ.
56

57. The first integral equals zero since the integrand is entire (a constant
function). For the second integral, we observe that
I
|ζ|=3
1
ζ−z
dζ=
(
0 if|z|>3
2πiif|z|<3
.
Furthermore, we deduce that the third integral is zero since
I
|ζ|=3
1
(ζ−z)
2
dζ=
d
dz
"
I
|ζ|=3
1
ζ−z
dζ
#
= 0.
Combining the results, we have
f
′
(z) = (2πi)(6z+7) if|z|<3.
We observe that 1 + iis inside|z|<3 since|1 +i|=
√
2<3.
Therefore, we obtain
f
′
(1+i) = 2πi[6(1+i)+7] =−12π+26πi.
57

58. Example
EvaluateI
C
e
2z
(z+1)
4
dz, whereCis the circle|z|= 3.
Solution
Letf(z) =e
2z
anda=−1 in the Cauchy integral formula
f
(n)
(a) =
n!
2πi
I
C
f(z)
(z−a)
n+1
dz.
Ifn= 3, thenf
′′′
(z) = 8e
2z
andf
′′′
(−1) = 8e
−2
. Hence,
8e
−2
=
3!
2πi
I
e
2z
(z+1)
4
dz
from which we see the required integral has the value 8πie
−2
/3.
58

59. Cauchy inequality
Supposef(z) is analytic on and inside the disc|z−z0|=r,0< r <∞,
and let
M(r) = max
|z−z
0
|=r
|f(z)|,
then
|f
(k)
(z)|
k!
≤
M(r)
r
k
, k= 0,1,2,....
This inequality follows from the generalized Cauchy integral formula.
59

60. Example
Supposef(z) is analytic inside the unit circle|z|= 1 and
|f(z)| ≤
1
1−|z|
,
show that
|f
(n)
(0)| ≤(n+1)!
θ
1+
1
n

n
.
Solution
We integrate
f(ζ)
ζ
n+1
around the circle|ζ|=
n
n+1
, wheref(ζ) is analytic
on and inside the circle. Using the generalized Cauchy integ ral
formula, we have
f
(n)
(0) =
n!
2πi
I
|ζ|=
n
n+1
f(ζ)ζ
n+1
dζ
60

61. =
n!
2πi
Z
2π
0
f
ζ
n
n+1
e
iθ

ζ
n
n+1

n+1
e
i(n+1)θ
θ
n
n+1

e
iθ
i dθ
=
θ
1+
1
n

n
n!
2π
Z
2π
0
f
θ
nn+1
e
iθ

e
−inθ
dθ.
The modulus|f
(n)
(0)|is bounded by
|f
(n)
(0)| ≤
θ
1+
1
n

n
n!
2π
Z
2π
0






f
θ
n
n+1
e
iθ







dθ
≤
θ
1+
1
n

n
n!
2π
Z
2π
0
1
1−
n
n+1
dθ
=
θ
1+
1
n

n
n!
2π
[(n+1) 2π]
= (n+1)!
θ
1+
1
n

n
.
61

62. Gauss’ mean value theorem
Iff(z)is analytic on and inside the discCr:|z−z0|=r, then
f(z0) =
1
2π
Z
2π
0
f(z0+re
iθ
)dθ.
Proof
From the Cauchy integral formula, we have
f(z0) =
1
2πi
I
C
r
f(z)
z−z0
dz
=
1
2πi
Z
2π
0
f(z0+re
iθ
)ire
iθ
re
iθ
dθ
=
1
2π
Z
2π
0
f(z0+re
iθ
)dθ.
Writeu(z) = Ref(z), it is known thatuis harmonic. We have
u(z0) =
1
2π
Z
2π
0
u(z0+re
iθ
)dθ.
62

63. Example
Find the mean value ofx
2
−y
2
+xon the circle|z−i|= 2.
Solution
First, we observe thatx
2
−y
2
+x= Re(z
2
+z). The mean value
is defined by
1
2π
Z
2π
0
u(i+2e
iθ
)dθ,
whereu(z) = Re(z
2
+z).By Gauss’ mean value theorem,
1
2π
Z
2π
0
u(i+2e
iθ
)dθ= Re(z
2
+z)






z=i
= Re(−1+i) =−1.
63

64. Maximum modulus theorem
Iff(z) is analytic on and inside a domainDbounded by a simple
closed curveC, then the maximum value of |f(z)|occurs onC,
unlessf(z) is a constant.
Example
Find the maximum value of |z
2
+3z−1|in the disk|z| ≤1.
Solution
The triangle inequality gives
|z
2
+3z−1| ≤ |z
2
|+3|z|+1≤5,for|z| ≤1.
64

65. However, the maximum value is actually smaller than this, as the
following analysis shows.
The maximum of|z
2
+3z−1|must occur on the boundary of the disk
(|z|= 1). The latter can be parameterized as z=e
it
,0≤t≤2π;
whence
|z
2
+3z−1|
2
= (e
i2t
+3e
it
−1)(e
−i2t
+3e
−it
−1).
Expanding and gathering terms reduces this to (11−2cos2t), whose
maximum value is 13. The maximum value is obtained by taking
t=π/2 ort= 3π/2.
Thus the maximum value of |z
2
+ 3z−1|is
√
13, which occurs at
z=±i.
65

66. Example
LetRdenote the rectangular region:
0≤x≤π,0≤y≤1,
the modulus of the entire function
f(z) = sinz
has a maximum value inRthat occurs on the boundary.
To verify the claim, consider
|f(z)|=
q
sin
2
x+sinh
2
y,
the term sin
2
xis greatest atx=π/2 and the increasing function
sinh
2
yis greatest wheny= 1. The maximum value of |f(z)|inR
occurs at the boundary point
θ
π
2
,1

and at no other point inR.
66

67. Proof of the Maximum Modulus Theorem
Proof by contradiction. Suppose|f(z)|attains its maximum atα∈
D. Using the Gauss Mean Value Theorem:
f(α) =
1
2π
Z
2π
0
f(α+re
iθ
)dθ
where the neighborhoodN(α;r) lies insideD. By the modulus in-
equality,
|f(α)| ≤
1
2π
Z
2π
0
|f(α+re
iθ
)|dθ.
Since|f(α)|is a maximum, then|f(α+re
iθ
)| ≤ |f(α)|for allθ, giving
1
2π
Z
2π
0
|f(α+re
iθ
)|dθ≤
1
2π
Z
2π
0
|f(α)|dθ=|f(α)|.
Combining the results, we obtain
Z
2π
0
[|f(α)|−|f(α+re
iθ
)|]
|
{z
}
non-negative
dθ= 0.
67

68. One then infer that|f(α)|=|f(α+re
iθ
)|. However, it may be
possible to have|f(α+re
iθ
)|<|f(α)|at isolated points. We argue
that this is not possible due to continuity off(z).
If|f(α+re
iθ
)|<|f(α)|at a single point, then|f(α+re
iθ
)|<|f(α)|
for a finite arc on the circle, giving
1
2π
Z
2π
0
|f(α+re
iθ
)|dθ <|f(α)|,
a contradiction. We can then deduce that
|f(α)|=|f(α+re
iθ
)|
for all points on the circle.
•Sincercan be any value,|f(z)|is constant in any neighborhood
ofαlying insideD.
68

69. Finally, we need to show that|f(z)|is constant at any point inD.
Take anyz∈ D, we can joinαtozby a curve lying completely
insideD. Taking a sequence of pointsz0=α,z1,∆∆∆,zn=zsuch
that each of these points is the center of a disc (plus its boundary)
lying completely insideDandz
kis contained in the disk centered at
z
k−1,k= 1,2,∆∆∆,n.
We then have|f(z1)|=|f(α)|. Alsoz2is contained inside the disc
centered atz1, so|f(z2)|=|f(z1)|,∆∆∆, and deductively|f(z)|=
|f(α)|.
Lastly, we use the result that if|f(z)|= constant throughout D,
thenf(z) = constant throughoutD.
69

70. Liouville’s Theorem
Iffis entire and bounded in the complex plane, thenf(z) is constant
throughout the plane.
Proof
It suffices to show thatf
′
(z) = 0 for allz∈C. We integrate
f(ζ)
(ζ−z)
2
aroundC
R:|ζ−z|=R. By the generalized Cauchy integral formula
f
′
(z) =
1
2πi
I
C
R
f(ζ)
(ζ−z)
2
dξ,
which remains valid for any sufficiently largeRsincef(z) is entire.
Sincef(z) is bounded, so|f(z)| ≤Bfor allz∈C,
|f
′
(z)|=
1
2π





I
C
R
f(ζ)
(ζ−z)
2
dζ





≤
1
2π
BR
2
2πR=
B
R
.
70

71. Now,Bis independent ofRandRcan be arbitrarily large. The
inequality can hold for arbitrarily large values ofRonly iff
′
(z) = 0.
Since the choice ofzis arbitrary, this means thatf
′
(z) = 0 every-
where in the complex plane. Consequently,fis a constant function.
Remark
Non-constant entire functions must be unbounded. For examp le,
sinzand coszare unbounded, unlike their real counterparts.
71