$\big| |z|-|w|\big| \leq |z-w| \implies \big|c_{1} |z|- |w|c_{2}\big| \leq |c_{1}z- c_{2} w| ? $ ($z, w \in \mathbb C, C_{1}, C_{2} >0$)

Question

By triangle inequality, we get, $$\big| |z|-|w|\big| \leq |z-w|; (z, w\in \mathbb C.)$$

Take any $C_{1}, C_{2} > 0$ and fix it.

My Question is: Can we expect: $$\big|C_{1} |z|- |w|C_{2}\big| \leq |C_{1}z- C_{2} w| ?$$

Thanks,

Answer 1

Of course. Take $z\mapsto C_1z$ and $w\mapsto C_2w$ in your original equation to get

$$\big||C_1z|-|C_2w|\big|\le\big|C_1z-C_2w\big|;$$

your equation arises because $C_1,C_2\ge0$ implies $|C_1z|=C_1|z|$ and $|C_2w|=C_2|w|$.

Answer 2

As, $C_1,C_2 > 0$ your claim is obvious.

$C_1|z| = |C_1z|$, as $C_1 > 0$. Let, $C_1z=x$ and $C_2w = y$. Then by simple algebra and triangle inequality, $|C_1|z|-|w|C_2|= ||C_1z|-|C_2w|=||x|-|y|| \leq |x-y| = |C_1z-C_2w|$