dx
+
@f
@y
dy
@f
@z
dz
+
@f
@
9
z
d
0
z
=
1
2
4
@f
@x
"
i
@f
@y
5
(
dx
+
idy
) +
1
2
4
@f
@x
+
i
@f
@y
5
(
dx
"
idy
) =
1
2
@f
@x
dx
+
1
2
@f
@y
dy
+
1
2
@f
@x
dx
+
1
2
@f
@y
dy
+
i
2
h
@f
@x
dy
"
@f
@y
dx
"
@f
@x
dy
+
@f
@y
dx
i
=
df
49

IV.8.6
1
2
3
P
L
K
Show that if
D
is a domain with smooth boundary, and if
f
(
z
)
and
g
(
z
)
are
analytic on
D
[
@D
, then
Z
@D
f
(
z
)
g
(
z
)
dz
= 2
i
ZZ
D
f
(
z
)
g
0
(
z
)
dx dy:
Compare this formula with Exercise 1.4.
Solution
By (8.4),
R
@D
f
(
z
)
g
(
z
)
dz
= 2
i
RR
D
@
@E
(
f
0
g
)
dz
= 2
i
RR
D
fgdz
Using the Leibnitz rule, obtain
@
@
9
z
(
f
0
g
) =
f
@
9
g
@
9
z
+ 0
g
@f
@
9
z
=
f
@g
@z
=
f
g
0
To get Ex1.4, take
f
= 1
,
g
(
z
) =
z
, get
R
@D
0
z dz
= 2
i
RR
D
dxdt
= 2
i
Area
50

IV.8.7
1
2
3
P
L
K
LLL
Show that the Taylor series expansion at
z
0
= 0
of a smooth function
f
(
z
)
, through the quadratic terms, is given by
f
(
z
) =
f
(0)+
@f
@z
(0)
z
+
@f
@
0
z
(0) 0
z
+
1
2
#
@
2
f
@z
2
(0)
z
2
+ 2
@
2
f
@[email protected]
0
z
(0)
j
z
j
2
$
+
o
%
j
z
j
3
&
:
Solution
Since a smooth function
f
is a linear combination of the functions
1
; z;
0
z; z
2
;
0
z
2
;
j
z
j
2
and a reminder term
O
%
j
z
j
3
&
, it su¢ces to check the formula for these six
functions. The formula holds fro each of these. For instance, for
j
z
j
2
=
z
0
z
are harmonic.
@f
@z
(0) = 0
,
@f
@
9
z
(0) = 0
,
@
2
f
@
9
z
2
(0) = 0
,
f
(0) = 0
,
@
2
f
@[email protected]
9
z
(0) = 1
.
The term
O
%
j
z
j
3
&
has Taylor series
0 +
O
%
j
z
j
3
&
. That ????? it.
51

IV.8.8
1
2
3
P
L
K
LLL
Establish the following version of the chain rule for smooth complex-
valued functions
w
=
w
(
z
)
and
h
=
h
(
w
)
.
@
@
0
z
(
h
7
w
)
=
@h
@w
@w
@
0
z
+
@h
@
0
w
@
0
w
@
0
z
;
@
@z
(
h
7
w
)
=
@h
@w
@w
@z
+
@h
@
0
w
@
0
w
@z
:
Solution
Suppose
w
(0) = 0
,
h
(0) = 0
, write
h
(
w
) =
aw
+
b
0
w
+
O
%
j
w
j
2
&
,
a
=
@h
@w
(0)
,
b
=
@h
@
9
w
(0)
and
w
(
z
) =
Yw
+
Z
0
w
+
O
%
j
z
j
2
&
,
Y
=
@w
@z
(0)
,
b
=
@w
@
9
z
(0)
then
h
(
w
(
z
)) =
a
%
Yz
+
Z
0
z
+
O
%
j
z
j
2
&&
+
b
%
0
Y
0
z
+
0
Zz
+
O
%
j
z
j
2
&&
+
O
%
j
w
j
2
&
=
aYz
+
aZ
0
z
+
b
0
Y
0
z
+
b
0
Zz
+
O
%
j
z
j
2
&
=
%
aY
+
b
0
Z
&
z
+ (
aZ
+
b
0
Y
) 0
z
+
O
%
j
z
j
2
&
This gives,
@h
,
w
@
9
z
(0) =
aZ
+
b
0
Y
=
@h
@w
@w
@
9
z
+
@h
@
9
w
@w
@z @
9
[email protected]
9
z
=
@h
@w
@w
@
9
z
+
@h
@
9
w
@
9
w
@
9
z
@h
,
w
@z
(0) =
aY
+
b
0
Z
=
@h
@w
@w
@z
+
@h
@
9
w
@w
@
0
z
|{z}
@
9
[email protected]
=
@h
@w
@w
@z
+
@h
@
9
w
@
9
w
@z
52

IV.8.9
1
2
3
P
L
K
LLL
Show with the aid of the preceding exercise that if both
h
(
w
)
and
w
(
z
)
are analytic, then
(
h
7
w
) (
z
)
and
(
h
7
w
)
0
(
z
) =
h
0
(
w
(
z
))
w
0
(
z
)
.
Solution
If the functions
h
and
w
are analytic, then
@h
,
w
@
9
z
= 0 + 0 = 0
so
h
7
w
is
analytic.
(
h
7
w
)
0
(
z
) =
@
@z
h
7
w
=
@h
@w
@w
@z
+ 0 =
h
0
(
w
(
z
))
w
0
(
z
)
53

IV.8.10
1
2
3
P
L
K
Let
g
(
z
)
be a continuously di§erentiable function on the complex plane that
is zero outside of som compact set. Show that
g
(
w
) =
"
1
;
ZZ
C
@g
@
0
z
1
z
"
w
dx dy;
w
2
C
:
Remark. If we integrate this formally by parts, we obtain
g
(
w
) =
1
;
ZZ
C
g
(
z
)
@
@
0
z
6
1
z
"
w
7
dx dy:
Thus the "distribution derivative" of
1
=
(
;
(
z
"
w
))
with respect to
z
is the
point mass
w
("Dirac delta-function"), in the sense that it is equal to
0
away
from
w
, and it is inÖnite at
w
in such a way that its integral (total mass) is
equal to 1.
Solution
Apply Pompeeiuës formula to a large disk
fj
z
j
< R
g
.
54

V
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
1
2
3
4
5
6
7
8
1

V.1.1
1
2
3
P
L
K
(Harmonic Series) Show that
n
X
k
=1
1
k
!
log
n:
Deduce that the series
P
1
k
=1
1
k
does not converge.
Hint
.
Use the
estimate
1
k
!
Z
k
+1
k
1
x
dx:
Solution
Let
k
be a nonnegative integer. Then since the function
x
7!
1
=x
is decreasing
on the positive reals, we have for all
x
2
[
k; k
+ 1]
that
1
=x
%
1
=k
. Hence
locking at the oversum in the interval
[
k; k
+ 1]
1
k
((
k
+ 1)
&
k
) =
1
k
!

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