Contour Integral, Cauchy’s Theorem, Cauchy’s Integral Formula - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


3 Contour integrals and Cauchy's Theorem
3.1 Line integrals of complex functions
Our goal here will be to discuss integration of complex functions f(z) =
u +iv, with particular regard to analytic functions. Of course, one way to
think of integration is as antidierentiation. But there is also the denite
integral. For a function f(x) of a real variable x, we have the integral
Z
b
a
f(x)dx. In case f(x) = u(x) +iv(x) is a complex-valued function of
a real variable x, the denite integral is the complex number obtained by
integrating the real and imaginary parts off(x) separately, i.e.
Z
b
a
f(x)dx =
Z
b
a
u(x)dx +i
Z
b
a
v(x)dx. For vector elds F = (P;Q) in the plane we have
the line integral
Z
C
Pdx+Qdy, whereC is an oriented curve. In caseP and
Q are complex-valued, in which case we call Pdx +Qdy a complex 1-form,
we again dene the line integral by integrating the real and imaginary parts
separately. Next we recall the basics of line integrals in the plane:
1. The vector eld F = (P;Q) is a gradient vector eldrg, which we
can write in terms of 1-forms as Pdx +Qdy = dg, if and only if
R
C
Pdx+Qdy only depends on the endpoints ofC, equivalently if and
only if
R
C
Pdx+Qdy = 0 for every closed curveC. IfPdx+Qdy =dg,
and C has endpoints z
0
and z
1
, then we have the formula
Z
C
Pdx +Qdy =
Z
C
dg =g(z
1
)g(z
0
):
2. If D is a plane region with oriented boundary @D =C, then
Z
C
Pdx +Qdy =
ZZ
D

@Q
@x

@P
@y

dxdy:
3. IfD is a simply connected plane region, then F = (P;Q) is a gradient
vector eldrg if and only if F satises the mixed partials condition
@Q
@x
=
@P
@y
.
(Recall that a region D is simply connected if every simple closed curve in
D is the boundary of a region contained in D. Thus a diskfz2C :jzj< 1g
Page 2


3 Contour integrals and Cauchy's Theorem
3.1 Line integrals of complex functions
Our goal here will be to discuss integration of complex functions f(z) =
u +iv, with particular regard to analytic functions. Of course, one way to
think of integration is as antidierentiation. But there is also the denite
integral. For a function f(x) of a real variable x, we have the integral
Z
b
a
f(x)dx. In case f(x) = u(x) +iv(x) is a complex-valued function of
a real variable x, the denite integral is the complex number obtained by
integrating the real and imaginary parts off(x) separately, i.e.
Z
b
a
f(x)dx =
Z
b
a
u(x)dx +i
Z
b
a
v(x)dx. For vector elds F = (P;Q) in the plane we have
the line integral
Z
C
Pdx+Qdy, whereC is an oriented curve. In caseP and
Q are complex-valued, in which case we call Pdx +Qdy a complex 1-form,
we again dene the line integral by integrating the real and imaginary parts
separately. Next we recall the basics of line integrals in the plane:
1. The vector eld F = (P;Q) is a gradient vector eldrg, which we
can write in terms of 1-forms as Pdx +Qdy = dg, if and only if
R
C
Pdx+Qdy only depends on the endpoints ofC, equivalently if and
only if
R
C
Pdx+Qdy = 0 for every closed curveC. IfPdx+Qdy =dg,
and C has endpoints z
0
and z
1
, then we have the formula
Z
C
Pdx +Qdy =
Z
C
dg =g(z
1
)g(z
0
):
2. If D is a plane region with oriented boundary @D =C, then
Z
C
Pdx +Qdy =
ZZ
D

@Q
@x

@P
@y

dxdy:
3. IfD is a simply connected plane region, then F = (P;Q) is a gradient
vector eldrg if and only if F satises the mixed partials condition
@Q
@x
=
@P
@y
.
(Recall that a region D is simply connected if every simple closed curve in
D is the boundary of a region contained in D. Thus a diskfz2C :jzj< 1g
is simply connected, whereas a \ring" such asfz2C : 1<jzj< 2g is not.)
In case Pdx +Qdy is a complex 1-form, all of the above still makes sense,
and in particular Green's theorem is still true.
We will be interested in the following integrals. Let dz = dx +idy, a
complex 1-form (with P = 1 and Q = i), and let f(z) = u +iv. The
expression
f(z)dz = (u +iv)(dx +idy) = (u +iv)dx + (iuv)dy
= (udxvdy) +i(vdx +udy)
is also a complex 1-form, of a very special type. Then we can dene
Z
C
f(z)dz for any reasonable closed oriented curveC. IfC is a parametrized
curve given by r(t), atb, then we can view r
0
(t) as a complex-valued
curve, and then
Z
C
f(z)dz =
Z
b
a
f(r(t))
r
0
(t)dt;
where the indicated multiplication is multiplication of complex numbers
(and not the dot product). Another notation which is frequently used is
the following. We denote a parametrized curve in the complex plane byz(t),
atb, and its derivative by z
0
(t). Then
Z
C
f(z)dz =
Z
b
a
f(z(t))z
0
(t)dt:
For example, let C be the curve parametrized by r(t) =t + 2t
2
i, 0t 1,
and let f(z) =z
2
. Then
Z
C
z
2
dz =
Z
1
0
(t + 2t
2
i)
2
(1 + 4ti)dt =
Z
1
0
(t
2
 4t
4
+ 4t
3
i)(1 + 4ti)dt
=
Z
1
0
[(t
2
 4t
4
 16t
4
) +i(4t
3
+ 4t
3
 16t
5
)]dt
=t
3
=3 4t
5
+i(2t
4
 8t
6
=3)]
1
0
=11=3 + (2=3)i:
For another example, let let C be the unit circle, which can be eciently
parametrized as r(t) = e
it
= cost +i sint, 0 t 2, and let f(z) =  z.
Then
r
0
(t) = sint +i cost =i(cost +i sint) =ie
it
:
Note that this is what we would get by the usual calculation of
d
dt
e
it
. Then
Z
C
 zdz =
Z
2
0
e
it

ie
it
dt =
Z
2
0
e
it

ie
it
dt =
Z
2
0
idt = 2i:
Page 3


3 Contour integrals and Cauchy's Theorem
3.1 Line integrals of complex functions
Our goal here will be to discuss integration of complex functions f(z) =
u +iv, with particular regard to analytic functions. Of course, one way to
think of integration is as antidierentiation. But there is also the denite
integral. For a function f(x) of a real variable x, we have the integral
Z
b
a
f(x)dx. In case f(x) = u(x) +iv(x) is a complex-valued function of
a real variable x, the denite integral is the complex number obtained by
integrating the real and imaginary parts off(x) separately, i.e.
Z
b
a
f(x)dx =
Z
b
a
u(x)dx +i
Z
b
a
v(x)dx. For vector elds F = (P;Q) in the plane we have
the line integral
Z
C
Pdx+Qdy, whereC is an oriented curve. In caseP and
Q are complex-valued, in which case we call Pdx +Qdy a complex 1-form,
we again dene the line integral by integrating the real and imaginary parts
separately. Next we recall the basics of line integrals in the plane:
1. The vector eld F = (P;Q) is a gradient vector eldrg, which we
can write in terms of 1-forms as Pdx +Qdy = dg, if and only if
R
C
Pdx+Qdy only depends on the endpoints ofC, equivalently if and
only if
R
C
Pdx+Qdy = 0 for every closed curveC. IfPdx+Qdy =dg,
and C has endpoints z
0
and z
1
, then we have the formula
Z
C
Pdx +Qdy =
Z
C
dg =g(z
1
)g(z
0
):
2. If D is a plane region with oriented boundary @D =C, then
Z
C
Pdx +Qdy =
ZZ
D

@Q
@x

@P
@y

dxdy:
3. IfD is a simply connected plane region, then F = (P;Q) is a gradient
vector eldrg if and only if F satises the mixed partials condition
@Q
@x
=
@P
@y
.
(Recall that a region D is simply connected if every simple closed curve in
D is the boundary of a region contained in D. Thus a diskfz2C :jzj< 1g
is simply connected, whereas a \ring" such asfz2C : 1<jzj< 2g is not.)
In case Pdx +Qdy is a complex 1-form, all of the above still makes sense,
and in particular Green's theorem is still true.
We will be interested in the following integrals. Let dz = dx +idy, a
complex 1-form (with P = 1 and Q = i), and let f(z) = u +iv. The
expression
f(z)dz = (u +iv)(dx +idy) = (u +iv)dx + (iuv)dy
= (udxvdy) +i(vdx +udy)
is also a complex 1-form, of a very special type. Then we can dene
Z
C
f(z)dz for any reasonable closed oriented curveC. IfC is a parametrized
curve given by r(t), atb, then we can view r
0
(t) as a complex-valued
curve, and then
Z
C
f(z)dz =
Z
b
a
f(r(t))
r
0
(t)dt;
where the indicated multiplication is multiplication of complex numbers
(and not the dot product). Another notation which is frequently used is
the following. We denote a parametrized curve in the complex plane byz(t),
atb, and its derivative by z
0
(t). Then
Z
C
f(z)dz =
Z
b
a
f(z(t))z
0
(t)dt:
For example, let C be the curve parametrized by r(t) =t + 2t
2
i, 0t 1,
and let f(z) =z
2
. Then
Z
C
z
2
dz =
Z
1
0
(t + 2t
2
i)
2
(1 + 4ti)dt =
Z
1
0
(t
2
 4t
4
+ 4t
3
i)(1 + 4ti)dt
=
Z
1
0
[(t
2
 4t
4
 16t
4
) +i(4t
3
+ 4t
3
 16t
5
)]dt
=t
3
=3 4t
5
+i(2t
4
 8t
6
=3)]
1
0
=11=3 + (2=3)i:
For another example, let let C be the unit circle, which can be eciently
parametrized as r(t) = e
it
= cost +i sint, 0 t 2, and let f(z) =  z.
Then
r
0
(t) = sint +i cost =i(cost +i sint) =ie
it
:
Note that this is what we would get by the usual calculation of
d
dt
e
it
. Then
Z
C
 zdz =
Z
2
0
e
it

ie
it
dt =
Z
2
0
e
it

ie
it
dt =
Z
2
0
idt = 2i:
One nal point in this section: let f(z) =u +iv be any complex valued
function. Then we can computerf, or equivalently df. This computation
is important, among other reasons, because of the chain rule: if r(t) =
(x(t);y(t)) is a parametrized curve in the plane, then
d
dt
f(r(t)) =rf 
r
0
(t) =
@f
@x
dx
dt
+
@f
@y
dy
dt
:
(Here 
 means the dot product.) We can think of obtaining
d
dt
f(r(t)) roughly
by taking the formal denition df =
@f
@x
dx +
@f
@y
dy and dividing both sides
by dt.
Of course we expect that df should have a particularly nice form if f(z)
is analytic. In fact, for a general function f(z) =u +iv, we have
df =

@u
@x
+i
@v
@x

dx +

@u
@y
+i
@v
@y

dy
and thus, if f(z) is analytic,
df =

@u
@x
+i
@v
@x

dx +


@v
@x
+i
@u
@x

dy
=

@u
@x
+i
@v
@x

dx +

@u
@x
+i
@v
@x

idy =

@u
@x
+i
@v
@x

(dx +idy) =f
0
(z)dz:
Hence: if f(z) is analytic, then
df =f
0
(z)dz
and thus, if z(t) = (x(t);y(t)) is a parametrized curve, then
d
dt
f(z(t)) =f
0
(z(t))z
0
(t)
This is sometimes called the chain rule for analytic functions. For example,
if  = a +bi is a complex number, then applying the chain rule to the
analytic function f(z) =e
z
and z(t) =t =at + (bt)i, we see that
d
dt
e
t
=e
t
:
Page 4


3 Contour integrals and Cauchy's Theorem
3.1 Line integrals of complex functions
Our goal here will be to discuss integration of complex functions f(z) =
u +iv, with particular regard to analytic functions. Of course, one way to
think of integration is as antidierentiation. But there is also the denite
integral. For a function f(x) of a real variable x, we have the integral
Z
b
a
f(x)dx. In case f(x) = u(x) +iv(x) is a complex-valued function of
a real variable x, the denite integral is the complex number obtained by
integrating the real and imaginary parts off(x) separately, i.e.
Z
b
a
f(x)dx =
Z
b
a
u(x)dx +i
Z
b
a
v(x)dx. For vector elds F = (P;Q) in the plane we have
the line integral
Z
C
Pdx+Qdy, whereC is an oriented curve. In caseP and
Q are complex-valued, in which case we call Pdx +Qdy a complex 1-form,
we again dene the line integral by integrating the real and imaginary parts
separately. Next we recall the basics of line integrals in the plane:
1. The vector eld F = (P;Q) is a gradient vector eldrg, which we
can write in terms of 1-forms as Pdx +Qdy = dg, if and only if
R
C
Pdx+Qdy only depends on the endpoints ofC, equivalently if and
only if
R
C
Pdx+Qdy = 0 for every closed curveC. IfPdx+Qdy =dg,
and C has endpoints z
0
and z
1
, then we have the formula
Z
C
Pdx +Qdy =
Z
C
dg =g(z
1
)g(z
0
):
2. If D is a plane region with oriented boundary @D =C, then
Z
C
Pdx +Qdy =
ZZ
D

@Q
@x

@P
@y

dxdy:
3. IfD is a simply connected plane region, then F = (P;Q) is a gradient
vector eldrg if and only if F satises the mixed partials condition
@Q
@x
=
@P
@y
.
(Recall that a region D is simply connected if every simple closed curve in
D is the boundary of a region contained in D. Thus a diskfz2C :jzj< 1g
is simply connected, whereas a \ring" such asfz2C : 1<jzj< 2g is not.)
In case Pdx +Qdy is a complex 1-form, all of the above still makes sense,
and in particular Green's theorem is still true.
We will be interested in the following integrals. Let dz = dx +idy, a
complex 1-form (with P = 1 and Q = i), and let f(z) = u +iv. The
expression
f(z)dz = (u +iv)(dx +idy) = (u +iv)dx + (iuv)dy
= (udxvdy) +i(vdx +udy)
is also a complex 1-form, of a very special type. Then we can dene
Z
C
f(z)dz for any reasonable closed oriented curveC. IfC is a parametrized
curve given by r(t), atb, then we can view r
0
(t) as a complex-valued
curve, and then
Z
C
f(z)dz =
Z
b
a
f(r(t))
r
0
(t)dt;
where the indicated multiplication is multiplication of complex numbers
(and not the dot product). Another notation which is frequently used is
the following. We denote a parametrized curve in the complex plane byz(t),
atb, and its derivative by z
0
(t). Then
Z
C
f(z)dz =
Z
b
a
f(z(t))z
0
(t)dt:
For example, let C be the curve parametrized by r(t) =t + 2t
2
i, 0t 1,
and let f(z) =z
2
. Then
Z
C
z
2
dz =
Z
1
0
(t + 2t
2
i)
2
(1 + 4ti)dt =
Z
1
0
(t
2
 4t
4
+ 4t
3
i)(1 + 4ti)dt
=
Z
1
0
[(t
2
 4t
4
 16t
4
) +i(4t
3
+ 4t
3
 16t
5
)]dt
=t
3
=3 4t
5
+i(2t
4
 8t
6
=3)]
1
0
=11=3 + (2=3)i:
For another example, let let C be the unit circle, which can be eciently
parametrized as r(t) = e
it
= cost +i sint, 0 t 2, and let f(z) =  z.
Then
r
0
(t) = sint +i cost =i(cost +i sint) =ie
it
:
Note that this is what we would get by the usual calculation of
d
dt
e
it
. Then
Z
C
 zdz =
Z
2
0
e
it

ie
it
dt =
Z
2
0
e
it

ie
it
dt =
Z
2
0
idt = 2i:
One nal point in this section: let f(z) =u +iv be any complex valued
function. Then we can computerf, or equivalently df. This computation
is important, among other reasons, because of the chain rule: if r(t) =
(x(t);y(t)) is a parametrized curve in the plane, then
d
dt
f(r(t)) =rf 
r
0
(t) =
@f
@x
dx
dt
+
@f
@y
dy
dt
:
(Here 
 means the dot product.) We can think of obtaining
d
dt
f(r(t)) roughly
by taking the formal denition df =
@f
@x
dx +
@f
@y
dy and dividing both sides
by dt.
Of course we expect that df should have a particularly nice form if f(z)
is analytic. In fact, for a general function f(z) =u +iv, we have
df =

@u
@x
+i
@v
@x

dx +

@u
@y
+i
@v
@y

dy
and thus, if f(z) is analytic,
df =

@u
@x
+i
@v
@x

dx +


@v
@x
+i
@u
@x

dy
=

@u
@x
+i
@v
@x

dx +

@u
@x
+i
@v
@x

idy =

@u
@x
+i
@v
@x

(dx +idy) =f
0
(z)dz:
Hence: if f(z) is analytic, then
df =f
0
(z)dz
and thus, if z(t) = (x(t);y(t)) is a parametrized curve, then
d
dt
f(z(t)) =f
0
(z(t))z
0
(t)
This is sometimes called the chain rule for analytic functions. For example,
if  = a +bi is a complex number, then applying the chain rule to the
analytic function f(z) =e
z
and z(t) =t =at + (bt)i, we see that
d
dt
e
t
=e
t
:
3.2 Cauchy's theorem
Suppose now thatC is a simple closed curve which is the boundary @D of a
region in C. We want to apply Green's theorem to the integral
Z
C
f(z)dz.
Working this out, since
f(z)dz = (u +iv)(dx +idy) = (udxvdy) +i(vdx +udy);
we see that
Z
C
f(z)dz =
ZZ
D


@v
@x

@u
@y

dA +i
ZZ
D

@u
@x

@v
@y

dA:
Thus, the integrand is always zero if and only if the following equations hold:
@v
@x
=
@u
@y
;
@u
@x
=
@v
@y
:
Of course, these are just the Cauchy-Riemann equations! This gives:
Theorem (Cauchy's integral theorem): LetC be a simple closed curve
which is the boundary@D of a region inC. Letf(z) be analytic inD. Then
Z
C
f(z)dz = 0:
Actually, there is a stronger result, which we shall prove in the next
section:
Theorem (Cauchy's integral theorem 2): LetD be a simply connected
region in C and let C be a closed curve (not necessarily simple) contained
in D. Let f(z) be analytic in D. Then
Z
C
f(z)dz = 0:
Example: let D =C and let f(z) be the function z
2
+z + 1. Let C be
the unit circle. Then as before we use the parametrization of the unit circle
given by r(t) =e
it
, 0t 2, and r
0
(t) =ie
it
. Thus
Z
C
f(z)dz =
Z
2
0
(e
2it
+e
it
+ 1)ie
it
dt =i
Z
2
0
(e
3it
+e
2it
+e
it
)dt:
Page 5


3 Contour integrals and Cauchy's Theorem
3.1 Line integrals of complex functions
Our goal here will be to discuss integration of complex functions f(z) =
u +iv, with particular regard to analytic functions. Of course, one way to
think of integration is as antidierentiation. But there is also the denite
integral. For a function f(x) of a real variable x, we have the integral
Z
b
a
f(x)dx. In case f(x) = u(x) +iv(x) is a complex-valued function of
a real variable x, the denite integral is the complex number obtained by
integrating the real and imaginary parts off(x) separately, i.e.
Z
b
a
f(x)dx =
Z
b
a
u(x)dx +i
Z
b
a
v(x)dx. For vector elds F = (P;Q) in the plane we have
the line integral
Z
C
Pdx+Qdy, whereC is an oriented curve. In caseP and
Q are complex-valued, in which case we call Pdx +Qdy a complex 1-form,
we again dene the line integral by integrating the real and imaginary parts
separately. Next we recall the basics of line integrals in the plane:
1. The vector eld F = (P;Q) is a gradient vector eldrg, which we
can write in terms of 1-forms as Pdx +Qdy = dg, if and only if
R
C
Pdx+Qdy only depends on the endpoints ofC, equivalently if and
only if
R
C
Pdx+Qdy = 0 for every closed curveC. IfPdx+Qdy =dg,
and C has endpoints z
0
and z
1
, then we have the formula
Z
C
Pdx +Qdy =
Z
C
dg =g(z
1
)g(z
0
):
2. If D is a plane region with oriented boundary @D =C, then
Z
C
Pdx +Qdy =
ZZ
D

@Q
@x

@P
@y

dxdy:
3. IfD is a simply connected plane region, then F = (P;Q) is a gradient
vector eldrg if and only if F satises the mixed partials condition
@Q
@x
=
@P
@y
.
(Recall that a region D is simply connected if every simple closed curve in
D is the boundary of a region contained in D. Thus a diskfz2C :jzj< 1g
is simply connected, whereas a \ring" such asfz2C : 1<jzj< 2g is not.)
In case Pdx +Qdy is a complex 1-form, all of the above still makes sense,
and in particular Green's theorem is still true.
We will be interested in the following integrals. Let dz = dx +idy, a
complex 1-form (with P = 1 and Q = i), and let f(z) = u +iv. The
expression
f(z)dz = (u +iv)(dx +idy) = (u +iv)dx + (iuv)dy
= (udxvdy) +i(vdx +udy)
is also a complex 1-form, of a very special type. Then we can dene
Z
C
f(z)dz for any reasonable closed oriented curveC. IfC is a parametrized
curve given by r(t), atb, then we can view r
0
(t) as a complex-valued
curve, and then
Z
C
f(z)dz =
Z
b
a
f(r(t))
r
0
(t)dt;
where the indicated multiplication is multiplication of complex numbers
(and not the dot product). Another notation which is frequently used is
the following. We denote a parametrized curve in the complex plane byz(t),
atb, and its derivative by z
0
(t). Then
Z
C
f(z)dz =
Z
b
a
f(z(t))z
0
(t)dt:
For example, let C be the curve parametrized by r(t) =t + 2t
2
i, 0t 1,
and let f(z) =z
2
. Then
Z
C
z
2
dz =
Z
1
0
(t + 2t
2
i)
2
(1 + 4ti)dt =
Z
1
0
(t
2
 4t
4
+ 4t
3
i)(1 + 4ti)dt
=
Z
1
0
[(t
2
 4t
4
 16t
4
) +i(4t
3
+ 4t
3
 16t
5
)]dt
=t
3
=3 4t
5
+i(2t
4
 8t
6
=3)]
1
0
=11=3 + (2=3)i:
For another example, let let C be the unit circle, which can be eciently
parametrized as r(t) = e
it
= cost +i sint, 0 t 2, and let f(z) =  z.
Then
r
0
(t) = sint +i cost =i(cost +i sint) =ie
it
:
Note that this is what we would get by the usual calculation of
d
dt
e
it
. Then
Z
C
 zdz =
Z
2
0
e
it

ie
it
dt =
Z
2
0
e
it

ie
it
dt =
Z
2
0
idt = 2i:
One nal point in this section: let f(z) =u +iv be any complex valued
function. Then we can computerf, or equivalently df. This computation
is important, among other reasons, because of the chain rule: if r(t) =
(x(t);y(t)) is a parametrized curve in the plane, then
d
dt
f(r(t)) =rf 
r
0
(t) =
@f
@x
dx
dt
+
@f
@y
dy
dt
:
(Here 
 means the dot product.) We can think of obtaining
d
dt
f(r(t)) roughly
by taking the formal denition df =
@f
@x
dx +
@f
@y
dy and dividing both sides
by dt.
Of course we expect that df should have a particularly nice form if f(z)
is analytic. In fact, for a general function f(z) =u +iv, we have
df =

@u
@x
+i
@v
@x

dx +

@u
@y
+i
@v
@y

dy
and thus, if f(z) is analytic,
df =

@u
@x
+i
@v
@x

dx +


@v
@x
+i
@u
@x

dy
=

@u
@x
+i
@v
@x

dx +

@u
@x
+i
@v
@x

idy =

@u
@x
+i
@v
@x

(dx +idy) =f
0
(z)dz:
Hence: if f(z) is analytic, then
df =f
0
(z)dz
and thus, if z(t) = (x(t);y(t)) is a parametrized curve, then
d
dt
f(z(t)) =f
0
(z(t))z
0
(t)
This is sometimes called the chain rule for analytic functions. For example,
if  = a +bi is a complex number, then applying the chain rule to the
analytic function f(z) =e
z
and z(t) =t =at + (bt)i, we see that
d
dt
e
t
=e
t
:
3.2 Cauchy's theorem
Suppose now thatC is a simple closed curve which is the boundary @D of a
region in C. We want to apply Green's theorem to the integral
Z
C
f(z)dz.
Working this out, since
f(z)dz = (u +iv)(dx +idy) = (udxvdy) +i(vdx +udy);
we see that
Z
C
f(z)dz =
ZZ
D


@v
@x

@u
@y

dA +i
ZZ
D

@u
@x

@v
@y

dA:
Thus, the integrand is always zero if and only if the following equations hold:
@v
@x
=
@u
@y
;
@u
@x
=
@v
@y
:
Of course, these are just the Cauchy-Riemann equations! This gives:
Theorem (Cauchy's integral theorem): LetC be a simple closed curve
which is the boundary@D of a region inC. Letf(z) be analytic inD. Then
Z
C
f(z)dz = 0:
Actually, there is a stronger result, which we shall prove in the next
section:
Theorem (Cauchy's integral theorem 2): LetD be a simply connected
region in C and let C be a closed curve (not necessarily simple) contained
in D. Let f(z) be analytic in D. Then
Z
C
f(z)dz = 0:
Example: let D =C and let f(z) be the function z
2
+z + 1. Let C be
the unit circle. Then as before we use the parametrization of the unit circle
given by r(t) =e
it
, 0t 2, and r
0
(t) =ie
it
. Thus
Z
C
f(z)dz =
Z
2
0
(e
2it
+e
it
+ 1)ie
it
dt =i
Z
2
0
(e
3it
+e
2it
+e
it
)dt:
It is easy to check directly that this integral is 0, for example because terms
such as
R
2
0
cos 3tdt (or the same integral with cos 3t replaced by sin 3t or
cos 2t, etc.) are all zero.
On the other hand, again with C the unit circle,
Z
C
1
z
dz =
Z
2
0
e
it
ie
it
dt =i
Z
2
0
dt = 2i6= 0:
The dierence is that 1=z is analytic in the regionCf0g =fz2C :z6= 0g,
but this region is not simply connected. (Why not?)
Actually, the converse to Cauchy's theorem is also true: if
Z
C
f(z)dz = 0
for every closed curve in a region D (simply connected or not), then f(z) is
analytic in D. We will see this later.
3.3 Antiderivatives
IfD is a simply connected region, C is a curve contained inD,P ,Q are de-
ned inD and
@Q
@x
=
@P
@y
, then the line integral
Z
C
Pdx+Qdy only depends
on the endpoints of C. However, if Pdx +Qdy =dF , then
Z
C
Pdx +Qdy
only depends on the endpoints of C whether or not D is simply connected.
We see what this condition means in terms of complex function theory: Let
f(z) =u +iv and suppose thatf(z)dz =dF , where we writeF in terms of
its real and imaginary parts as F =U +iV . This says that
(udxvdy) +i(vdx +udy) =

@U
@x
dx +
@U
@y
dy

+i

@V
@x
dx +
@V
@y
dy

:
Equating terms, this says that
u =
@U
@x
=
@V
@y
v =
@U
@y
=
@V
@x
:
In particular, we see that F satises the Cauchy-Riemann equations, and
its complex derivative is
F
0
(z) =
@U
@x
+i
@V
@x
=u +iv =f(z):