Can we deduce that $|z|≥|a|-|b|$

Question

Let us consider three complex numbers $z,a,b$ such that the equality $z=a-b$ holds true. Can we deduce that $$|z|≥|a|-|b|$$

Since |.| is a norm, the inequality holds (can be proved using triangle inequality). — Emo, Feb 20 '14 at 17:21

score 3 · Answer 1 · answered Feb 20 '14 at 17:20

3

Yes since by triangle inequality we have

$$|a-b|\ge|a|-|b|$$

answered Feb 20 '14 at 17:20

score 3 · Accepted Answer · answered Feb 20 '14 at 17:20

3

Yes.

$$ a = b + z $$ so $$ |a| = |b + z| \leq |b| + |z|. $$

In fact, even more is true! Since $b = a - z$, you can repeat this, $$ |b| = |a - z| \leq |a| + |z| $$

and combining these two, you have

$$ |z| \geq ||a| - |b|| $$

answered Feb 20 '14 at 17:20

BaronVT

score 1 · Answer 3 · answered Feb 20 '14 at 17:21

1

Yes, first $z=a-b\iff z+b=a$. Then

$$|a|=|z+b|\le |z|+|b|$$

Then subtract $|b|$ from both sides.

answered Feb 20 '14 at 17:21

David P

3 Answers3