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Math 5615H: Introduction to Analysis I.
Fall 2011
Homework #3. Problems and Solutions.
#1.
Let
A
:=
{
a
1
,a
2
,...
}
be a set of real numbers deﬁned as follows:
a
1
= 1
,
and
a
k
+1
= 1 +
√
a
k
for
k
= 1
,
2
,....
Find sup
A
.
Solution.
By induction, 1
≤
a
k
<
4 for all
k
≥
1, therefore,
∃
L
:= sup
A
∈
[1
,
4]. Since
a
k
+1
= 1 +
√
a
k
≤
1 +
√
L
for all
k
, we have
L
= sup
{
a
k
+1
} ≤
1 +
√
L
. On the other hand, from 1 +
√
a
k
=
a
k
+1
≤
L
it follows
1 +
√
L
≤
L
, because in an equivalent form,
a
k
= (
a
k
+1

1)
2
≤
(
L

1)
2
implies
L
:= sup
{
a
k
} ≤
(
L

1)
2
.
Therefore, 1 +
√
L
=
L
∈
[1
,
4], and
L
= (3 +
√
5)
/
2.
#2.
Let
z
1
and
z
2
be complex imaginary (not real) numbers such that both
z
1
+
z
2
and
z
1
z
2
are
real
. Show
that
z
2
=
z
1
, i.e.
z
2
is the
conjugate
of
z
1
.
Proof.
Denote
z
1
:=
a
1
+
b
1
i, z
2
=
a
2
+
b
2
i
. Since
z
1
+
z
2
= (
a
1
+
a
2
) + (
b
1
+
b
2
)
i
is real, we have
b
2
=

b
1
6
= 0. Since
z
1
z
2
is real, we also have Im(
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