Complex Analysis
Fall 2007
Homework 4: Solutions
1.5.2.
(a) The function
f
(
z
) = 3
z
2
+7
z
+5 is a polynomial so is analytic everywhere with derivative
f
(
z
) = 6
z
+ 7.
(b) The function
f
(
z
) = (2
z
+3)
4
is a composition of polynomials so is analytic everywhere
with derivative
f
(
z
) = 8(2
z
+ 3)
3
(by the chain rule).
(c) The function
f
(
z
) = (3
z

1)
/
(3

z
) is rational and so is analytic where
z
= 3. By the
quotient rule, its derivative is
f
(
z
) =
8
(3

z
)
2
.
1.5.10.
Let
f
(
z
) =

z

and write
f
=
u
+
iv
and
z
=
x
+
iy
. Then
u
=
x
2
+
y
2
and
v
= 0.
We find that
∂u
∂x
=
x
x
2
+
y
2
,
∂u
∂y
=
y
x
2
+
y
2
∂v
∂x
=
∂v
∂y
= 0
.
In order for
f
to be analytic the CauchyRiemann equations must hold. That is, we need
x
x
2
+
y
2
=
y
x
2
+
y
2
= 0
which is impossible, since it requires
x
and
y
to be both simultaneously zero and nonzero.
Therefore
f
is differentiable nowhere.
Here’s another proof.
Since the partial derivatives of Re(

z

) =
x
2
+
y
2
do not exist
at (0
,
0), the CauchyRiemann theorem implies that

z

cannot be differentiable at
z
= 0.
Suppose that
f
(
z
) =

z

were analytic at some
z
0
∈
C
,
z
0
= 0. Then (
f
(
z
))
2
=

z

2
=
z
z
would be analytic at
z
0
and, since
z
0
is nonzero, so too would be (
f
(
z
))
2
/z
=
z
. But we
know that
g
(
z
) =
z
is analytic nowhere, so this is impossible. Therefore
f
(
z
) =

z

cannot
be analytic at any nonzero complex number either. Hence,

z

is differentiable nowhere, i.e.
is not analytic.
1.5.14.
(a) According to CauchyRiemann equations, if
f
is analytic then
f
=
∂f
∂x
=
1
i
∂f
∂y
1
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so that
∂f
∂z
=
1
2
∂f
∂x
+
1
i
∂f
∂y
=
1
2
(
f
+
f
) =
f .
(b) If
f
(
z
) =
z
=
x
+
iy
then
f
is analytic so that by part (a) we have
∂fz
=
f
= 1
.
Moreover
∂f
∂x
= 1,
∂f
∂y
=
i
so that
∂f
∂
z
=
1
2
∂f
∂x

1
i
f
y
=
1
2
(1

1) = 0
.
(c) If
f
(
z
) =
z
=
x

iy
then
∂f
∂x
= 1,
∂f
∂y
=

i.
Therefore
∂f
∂z
=
1
2
∂f
∂x
+
1
i
∂f
∂y
=
1
2
(1

1) = 0
and
∂f
∂
z
=
1
2
∂f
∂x

1
i
∂f
∂y
=
1
2
(1 + 1) = 1
.
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 Fall '11
 RyanDaileda
 Polynomials, Derivative, ∂x

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