The Real And Complex Number Systems

so, |a − b| 2 ≤ ( 1 + |a| 2) ( 1 + |b| 2)
(b) If a ≠ 0, then |a + b| = |a| + |b| if, and only if, b/a is real and
nonnegative.
Proof: (⇒)Since |a + b| = |a| + |b| , we have
which implies that
which implies that
which implies that
Then
(⇐) Suppose that
|a + b| 2 = (|a| + |b|) 2
Re (āb) = |a| |b| = |ā| |b|
b
a = āb
āa
b
a
āb = |ā| |b|
=
|ā| |b|
|a| 2 ≥ 0.
= k, where k ≥ 0.
|a + b| = |a + ka| = (1 + k) |a| = |a| + k |a| = |a| + |b| .
1.33 If a and b are complex numbers, prove that
|a − b| = |1 − āb|
if, and only if, |a| = 1 or |b| = 1. For which a and b is the inequality
|a − b| < |1 − āb| valid
Proof: (⇔) Since
|a − b| = |1 − āb|
⇔ ( ā − ¯b ) (a − b) = (1 − āb) ( 1 − a¯b )
⇔ |a| 2 + |b| 2 = 1 + |a| 2 |b| 2
⇔ ( |a| 2 − 1 ) ( |b| 2 − 1 ) = 0
⇔ |a| 2 = 1 or |b| 2 = 1.
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