Triangle Inequality Proof: Show that:$||z_1|-|z_2||\leq |z_1+z_2|$

Question

I've been puzzled by this deceptively simple-looking complex number modulus inequality proof.

Show that:$$||z_1|-|z_2||\leq |z_1+z_2|.$$ State the condition for the equality to hold.

I tried

$\because|z_1|\leq |z_1+z_2|+|z_2|$, by triangle inequality

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

but then I do not factor the modulus on LHS and my proof is insufficient. I thought I could simply take the modulus both sides, but then I realised that for say $5>-6$, applying modulus in this manner yields $5>6$ which is obviously wrong.

I feel like perhaps I need to express $z_1$ and $z_2$ as $r_1cis\theta_1$ , etc but I am unsure in how to approach this method.

Any help would be greatly appreciated. Thanks in advance.

Answer 1

Hint:

Just exchange the roles of $z_1$ and $z_2$ and then use

• $\pm x \leq y \Leftrightarrow |x| \leq y$ for reals $x,y$

Answer 2

You have $|z_1|-|z_2|\leq |z_1-z_2|$

By similar argument switching $z_1$ and $z_2$ we get $|z_2|-|z_1|\leq |z_1-z_2|$

But $||z_1|-|z_2||$ is either $|z_1|-|z_2|$ if $|z_1|\geq |z_2|$ or $|z_2|-|z_1|$ if not.

Answer 3

The inequality is always true. We have $z_1+z_2-z_2=z_1\implies \mid z_1\mid\le\mid z_1+z_2\mid+\mid z_2\mid\implies \mid z_1\mid-\mid z_2\mid\le\mid z_1+z_2\mid$.

Similarly with $z_1$ and $z_2$ interchanged.

Thus $\mid\mid z_1\mid-\mid z_2\mid\mid\le\mid z_1+z_2\mid$.