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01.08 (1 +
i
)
3
= 1 + 3
i
+ 3
i
2
+
i
3
= 1 + 3
i

3

i
=

2 + 2
i
01.17
z
4
=

1 =
e
i
π
. So
z
=
e
i
(
π/
4+2
kπ/
4)
for
k
= 0
,
1
,
2
,
3. Note that
e
i
θ
= cos
θ
+
i
sin
θ
01.38
e
i
a
e
i
b
=
e
i
(
a
+
b
)
. By Euler’s formula,
(cos
a
+
i
sin
a
)(cos
b
+
i
sin
b
) = cos(
a
+
b
) +
i
sin(
a
+
b
)
That is,
(cos
a
cos
b

sin
a
sin
b
) +
i
(sin
a
cos
b
+ cos
a
sin
b
)
So we get
cos(
a
+
b
) = cos
a
cos
b

sin
a
sin
b,
sin(
a
+
b
) = sin
a
cos
b
+cos
a
sin
b.
02.03 (
b
*
d
)
*
c
=
e
*
c
=
a
and
b
*
(
d
*
c
) =
b
*
b
=
c
. So
*
is not associative.
02.04 The operator
*
is commutative since the table is symmetric.
02.08 First,
a
*
b
=
ab
+ 1 =
b
*
a
, So
*
is commutative. Second, (
a
*
b
)
*
c
=
(
ab
+ 1)
*
c
= (
ab
+ 1)
c
+ 1 =
abc
+
c
+ 1 and
a
*
(
b
*
c
) =
a
*
(
bc
+ 1) =
a
(
bc
+ 1) + 1 =
abc
+
a
+ 1. So
*
is not associative.
02.26 If
*
is associative and commutative, then
(
a
*
b
)
*
(
c
*
d
) = (
c
*
d
)
*
(
a
*
b
) = (
d
*
c
)
*
(
a
*
b
) = [(
d
*
c
)
*
a
]
*
b
02.36 Let
*
be a associative binary operation on
S
and
H
=
{
a
∈
S

a
*
x
=
x
*
a
for all
a
∈
S
}
. We show that
H
is closed under
*
. If
a, b
∈
H
,
then
a
*
x
=
x
*
a
and
b
*
x
=
x
*
b
for all
x
∈
S
. Now
(
a
*
b
)
*
x
=
a
*
(
b
*
x
) =
a
*
(
x
*
b
) = (
a
*
x
)
*
b
= (
x
*
a
)
*
b
=
x
*
(
a
*
b
)
Hence
a
*
b
∈
H
. This proves that
H
is closed under
*
.
03.03 The map
φ
(
n
) = 2
n
is not an isomorphism from
h
Z
,
+
i
to
h
Z
,
+
i
, since
it is not onto. For example, there is no integer
n
such that
φ
(
n
) = 1.
03.04 The map
φ
(
n
) =
n
+ 1 is not an isomorphism from
h
Z
,
+
i
to
h
Z
,
+
i
,
since it does not satisfy the homomorphism property:
φ
(
n
+
m
) =
n
+
m
+ 1 but
φ
(
n
) +
φ
(
m
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This note was uploaded on 10/12/2010 for the course MATH 5310 taught by Professor Staff during the Spring '08 term at Auburn University.
 Spring '08
 Staff
 Algebra

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