50
Lecture 7
f
′
(
z
) =
−
2(
z
4
+ 5
z
2
+ 12)
/
[
z
2
(
z
2
+ 4)
2
]
,
analytic in
C
−
{
0
,
±
2
i
}
.
(e).
f
is
di
ff
erentiable everywhere,
f
′
(
z
) =
−
2
ze
−
z
2
,
analytic everywhere (entire).
7.5.
a
= 1
, b
=
−
3
, c
=
−
3
, y
= 0 gives
f
(
x
) =
ix
3
,
and hence
f
(
z
) =
iz
3
and
dw/dz
= 3
iz
2
.
7.6.
u
x
= 2
x, u
y
= 2
y
⇒
v
y
= 2
x
and
v
x
=
−
2
y
⇒
v
= 2
xy
+
f
(
y
) and
v
=
−
2
xy
+
g
(
x
)
,
which is impossible.
7.7.
LHS = 2(
u
2
x
+
u
2
y
) + 2(
v
2
x
+
v
2
y
)
.
7.8.
(a).
u
(
x, y
) =
(
x
3
−
3
xy
2
)
/
(
x
2
+
y
2
)
,
(
x, y
)
̸
= (0
,
0)
0
,
(
x, y
) = (0
,
0)
and
v
(
x, y
) =
(
y
3
−
3
x
2
y
)
/
(
x
2
+
y
2
)
,
(
x, y
)
̸
= (0
,
0)
0
,
(
x, y
) = (0
,
0)
.
u
x
(0
,
0) = 1
, u
y
(0
,
0) = 0
,
v
x
(0
,
0) = 0
, v
y
(0
,
0) = 1
.
(b). For
z
̸
= 0
,
(
f
(
z
)
−
f
(0))
/
(
z
−
0) = (
z/z
)
2
.
Now see Problem 5.6 (a).
7.9.
Use rules for di
ff
erentiation.
7.10.
Since
u
r
=
u
x
x
r
+
u
y
y
r
, u
θ
=
u
x
x
θ
+
u
y
y
θ
,
we have
u
r
=
u
x
cos
θ
+
u
y
sin
θ
, u
θ
=
−
u
x
r
sin
θ
+
u
y
r
cos
θ
(7.9)
and
v
r
=
v
x
cos
θ
+
v
y
sin
θ
, v
θ
=
−
v
x
r
sin
θ
+
v
y
r
cos
θ
,
which in view of (6.5) is the same as
v
r
=
−
u
y
cos
θ
+
u
x
sin
θ
, v
θ
=
u
y
r
sin
θ
+
u
x
r
cos
θ
.
(7.10)
From (7.9) and (7.10), the CauchyRiemann conditions (7.7) are immediate.
Now, since
f
′
(
z
) =
u
x
+
iv
x
and
u
x
=
u
r
cos
θ
−
u
θ
sin
θ
r
=
u
r
cos
θ
+
v
r
sin
θ
,
v
x
=
v
r
cos
θ
−
v
θ
sin
θ
r
=
v
r
cos
θ
−
u
r
sin
θ
,
it follows that
f
′
(
z
) =
u
r
(cos
θ
−
i
sin
θ
) +
iv
r
(cos
θ
−
i
sin
θ
) =
e
−
i
θ
(
u
r
+
iv
r
)
.
Since
u
=
√
r
cos
θ
2
, v
=
√
r
sin
θ
2
, u
r
=
1
2
√
r
cos
θ
2
, u
θ
=
−
1
2
√
r
sin
θ
2
, v
r
=
1
2
√
r
sin
θ
2
, v
θ
=
1
2
√
r
cos
θ
2
,
and hence conditions (7.7) are satisfied. Thus,
f
is di
ff
erentiable at all
z
except
z
= 0
.
Furthermore, from (7.8) it follows
that
f
′
(
z
) =
e
−
i
θ
1
2
√
r
(cos
θ
2
+
i
sin
θ
2
) =
1
2
√
z
.
7.11.
Since
x
= (
z
+
z
)
/
2 and
y
= (
z
−
z
)
/
2
i, u
and
v
may be regarded as
functions of
z
and
z.
Thus, condition
w
z
= 0 is the same as
∂
u
∂
x
∂
x
∂
z
+
∂
u
∂
y
∂
y
∂
z
+
i
∂
v
∂
x
∂
x
∂
z
+
∂
v
∂
y
∂
y
∂
z
= 0
,
which is the same as
1
2
∂
u
∂
x
−
1
2
i
∂
u
∂
y
+
i
2
∂
v
∂
x
−
1
2
∂
v
∂
y
= 0
.
Now compare the real and imaginary parts.
7.12.
Let
f
=
u
+
iv.
From calculus, it su
ﬃ
ces to show that Jacobian
J
(
x
0
, y
0
) =
u
x
v
x
u
y
v
y
(
x
0
, y
0
)
̸
= 0
.
Now use (6.5) and (6.3) to obtain
J
(
x
0
, y
0
) =
u
2
x
(
x
0
, y
0
) +
v
2
x
(
x
0
, y
0
) =

f
′
(
z
0
)

2
̸
= 0
.
7.13.
(a).
i/
2
,
(b).
−
18
,
(c).
−
16
.
7.14.
f
=
u
+
iv, u
=
c
⇒
u
x
=
u
y
= 0
⇒
v
x
=
v
y
= 0
⇒
v
is also a
constant, and hence
f
is a constant.
7.15.
If
f
=
u
+
iv,
then
u
= 2
h
3
and
v
=
h.
Now, by the CauchyRiemann
conditions, we have 6
h
2
h
x
=
h
y
and
−
6
h
2
h
y
=
h
x
.
Thus,
−
12
h
4
h
x
=
h
x
,
or
h
x
(12
h
4
+ 1) = 0
,
which implies that
h
x
= 0
,
and from this
h
y
= 0
.