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16
1.35
Let
ε
1
,...,ε
n
be complex numbers with

ε
j
=
1 for all
j
, where
n
≥
2.
(i)
Prove that
±
±
±
n
X
j
=
1
ε
j
±
±
±
≤
n
X
j
=
1
±
±
ε
j
±
±
=
n
.
Solution.
The triangle inequality gives

u
+
v
≤
u
+
v

for all
complex numbers
u
and
v
, with no restriction on their norms. The
inductive proof is routine.
(ii)
Prove that there is equality,
±
±
±
n
X
j
=
1
ε
j
±
±
±
=
n
,
if and only if all the
ε
j
are equal.
Solution.
The proof is by induction on
n
≥
2.
For the base step, suppose that

ε
1
+
ε
2
=
2. Therefore,
4
=
ε
1
+
ε
2

2
=
(ε
1
+
ε
2
)
·
(ε
1
+
ε
2
)
=
ε
1

2
+
2
ε
1
·
ε
2
+
ε
2

2
=
2
+
2
ε
1
·
ε
2
.
Therefore, 2
=
1
+
ε
1
·
ε
2
, so that
1
=
ε
1
·
ε
2
=
ε
1

ε
2

cos
θ
=
cos
θ,
where
θ
is the angle between
ε
1
and
ε
2
(for

ε
1
=
1
=
ε
2

).
Therefore,
θ
=
0or
θ
=
π
, so that
ε
2
=±
ε
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 Fall '11
 KeithCornell

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