How to show that $|z_{1} + z_{2}| = |z_{1}| + |z_{2}| \implies |z_{1} - z_{2}| = ||z_{1}| - |z_{2}|| $

Question

I have to show that the following are equivalent:

$a)\arg(z_{1}) = \arg(z_{2}) \pm 2\pi n , (n\in \mathbb{Z})$

$b) \Re(z_{1}\bar{z_{2}}) = |z_{1}||z_{2}| $

$c)|z_{1} + z_{2}| = |z_{1}| + |z_{2}|$

$d) |z_{1} - z_{2}| = ||z_{1}| - |z_{2}||$

So far I've managed to prove $(a) \implies (b) $, $(b) \implies (c)$, and $(d) \implies (a)$.

All I can think of is trying to get from one expression to the other, but I haven't gotten it right: $$ |z_{1} - z_{2}| = |z_{1} + -(z_{2})| = |z_{1}| + |z_{2}| = - (-|z_{1}| - |z_{1}|) = |-|z_{1}| - |z_{2}|| = ||z_{1}| + |z_{2}||$$

Thanks.

Answer 1

You have $$(|z_1|+|z_2|)^2=|z_1+z_2|^2=|z_1|^2+|z_2|^2+2\mathrm{Re}(z_1\overline{z_2})\leqslant(|z_1|+|z_2|)^2$$ since $\mathrm{Re}(z)\leqslant|z|$ for all $z\in\mathbb{C}$ with equality iff $z\in\mathbb{R}^+$. Hence $z_1\overline{z_2}\in\mathbb{R}^+$. If $z_2=0$, it's ok, let us suppose $z_2\neq 0$ and let $\lambda\in\mathbb{R}^+$ such that $z_1=\lambda z_2$, we finally have $$ |z_1-z_2|=|z_1-\lambda z_1|=|z_1||1-\lambda|=|z_1||1-|\lambda||=||z_1|-|z_2|| $$

Answer 2

If $z_1 = a+bi$ and $z_2 = c + di$ then

$|z_1 + z_2| = |(a+b) + (c+d)i| = \sqrt{(a+b)^2 + (c+d)^2}$ and

$|z_1 + z_2|^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + cd)=|z_1|^2 + |z_2|^2 + 2(ab+cd)$.

And $(|z_1| + |z_2|)^2 = |z_1|^2 + 2|z_1z_2| + |z_2|^2$

So $(|z_1 + z_2|)= |z_1| + |z_2| \implies (ab+cd) = |z_1z_2| \implies$

$|z_1 - z_2|^2 = $

$a^2 + b^2 + c^2 - d^2 + 2(ab + cd)|z_1|^2 + |z_2|^2 - 2(ab+cd)=$

$|z_1|^2 - 2|z_1z_2| + |z_2|^2=$

$(|z_1| - |z_2|)^2$.

In fact $|z_1 + z_2| = |z_1| + |z_2| \iff$

$|z_1 + z_2|^2 = (|z_1| +|z_2|)^2 \iff$

$(ab+cd) = |z_1z_2| \iff$

$|z_1 - z_2|^2 = (|z_1| - |z_2|)^2 \iff$

$|z_1-z_2| = ||z_1| - |z_2||$.

Answer 3

Suppose $z_1 \neq 0$ otherwise the result is immediate. Divide by $z_1$ to get $|1+z| = 1+|z|$, where $z={z_2\over z_1}$.

Squaring gives $(1+z)(1+\overline{z})= 1 + |z|^2 + 2 \operatorname{re} z = 1+|z|^2+2 |z|$, from which we get $\operatorname{re} z = |z|$, and hence $z=|z|$, in particular, $z$ is real and positive. Hence $z_2=r z_1$ for some $r >0$.

Then $|z_1-z_2| = |1-r||z_1| = |z_1||1-r| = ||z_1|-|z_2||$.