0
$\begingroup$

Is the inequality $\lvert z_1 + z_2 \rvert \ge \lvert z_1 \rvert - \lvert z_2 \rvert$ incorrect, where $z_1$ and $z_2$ are any two complex numbers? I need an example to prove that it is. And in case it is correct, can you please give the proof? Thanks for any help.

3 Answers 3

1

We know $\forall z_1,z_2 \in C:|z_1+z_2|\leq |z_1|+|z_2|$ (triangle inequality)

put $z_1=z_1-z_2 $ $$|(z_1-z_2)+z_2|\leq |(z_1-z_2)|+|z_2|\\|z_1|\leq |z_1-z_2|+|z_2| \to \\ |z_1-z_2|\geq |z_1|-|z_2|$$ Use this fact $ |a||b|=|ab| \to |z_2|=(1)|z_2|=|-1||z_2|=|-z_2|$ so $$|z_1-(-z_2)|\geq |z_1|-|-z_2| \\\to \\ |z_1+z_2|\geq |z_1|-|z_2|$$

  • 0
    I guess $|(z_1-z_2)+z_2|\leq |(z_1-z_2)|+|z_2|$ implies $|z_1|\leq |z_1-z_2|+|z_2|$ and not $|-z_1|\leq |z_1-z_2|+|z_2|$. Am I correct? – 2017-02-27
  • 0
    yes .you are right – 2017-02-27
  • 0
    I hope I didn't do anything wrong by editing. – 2017-03-05
2

This inequality is always true. It follows easily by the triangle inequality.

Start with $$\vert z_1\vert\le\bigl\vert z_1-(-z_2)\bigr\vert+\left\vert-z_2\right\vert\;.$$

  • 0
    @Philippe Malot, thanks for changing "a" with "the". But when you edit LaTeX, please don't touch the maths. I wanted to have my post exactly in a form I wrote. I am its author, here you are the corrector only. – 2017-02-25
1

It's correct. From the inequality: $$||z_1|-|z_2||\leq |z_1-z_2|$$ if you substitute $z_2=-z_2$, you get: $$||z_1|-|-z_2||\leq |z_1-(-z_2)|=|z_1+z_2|$$

and now, recall that $a\leq|a|$ for every real number $a$ to conclude:

$$|z_1|-|z_2|\leq||z_1|-|z_2||\leq |z_1+z_2|$$