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Even after re-reading posts If $z,w\in \mathbb{C}^{\times}$, then $|z+w|=|z|+|w|$ iff $w=tz$ for some $t\gt 0$ and $|z+w|=|z|+|w|$ iff $z=cw$ a dozen times, I still don't understand how the proof for the implication $|z + w| = |z| + |w| \implies w = tz, t > 0$ works. Specifically, in $|z+w|=|z|+|w|$ iff $z=cw$ it is shown that $|z + w| = |z| + |w| \Longleftrightarrow \mathrm{Re}(z\cdot \bar{w}) = |z|\cdot |w|$, and in If $z,w\in \mathbb{C}^{\times}$, then $|z+w|=|z|+|w|$ iff $w=tz$ for some $t\gt 0$ the argument rests on using the exact same equality and mentioned above, but i.) dividing by the RHS ii.) writing $z/|z| = e^{i\theta}, w/|w| = e^{i\theta}$.

In both of these approaches it is unclear to me how we can state anything about the relationship of $w$ to $z$ by inspecting the real part of the product $z\cdot \bar{w}$, i.e. how can the $\mathrm{Re}(z\cdot \bar{w})$ be manipulated any further?

Edit: My bad, an additional condition is that $z \neq 0, w \neq 0$.

Epsilon Away
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$$ \operatorname{Re}(z\cdot \bar{w}) = |z|\cdot |w| = |z\cdot \bar{w}| $$ implies that $$ t = z\cdot \bar{w} $$ is a real and non-negative number, and so is (assuming $w \ne 0$) $$ \frac{z}{w} = \frac{z\cdot \bar{w}}{w\cdot \bar{w}} = \frac{t}{|w|^2} $$

Martin R
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  • I don't want to remove the acceptance, but I have to ask for further clarification on why $t = z \cdot \bar{w}$ is real based on $\mathrm{Re}(z \cdot \bar{w}) = |z|\cdot |w| = |z \cdot \bar{w}|$? I understand the the first equality in your response implies that the real part of $|z \cdot \bar{w}|$ is non-negative, but why would this show that the imaginary part is zero? – Epsilon Away Jul 18 '21 at 11:15
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    @EpsilonAway: If a complex number $\zeta = a+ib$ satisfies $\operatorname{Re}(\zeta) = \zeta$ then $a = \sqrt{a^2+b^2}$, and that is only possible if $b=0$ and $a \ge 0$. – Martin R Jul 18 '21 at 11:47
  • Argh, of course! From $\mathrm{Re}(z\cdot \bar{w}) = |z \cdot \bar{w}| = \sqrt{\mathrm{Re}(z \cdot \bar{w})^2 + \mathrm{Im}(z \cdot \bar{w})^2}$ we get necessarily that $Im(z \cdot \bar{w}) = 0$ – Epsilon Away Jul 18 '21 at 11:53
  • Exactly. (And I meant $\operatorname{Re}(\zeta) = |\zeta|$.) – Martin R Jul 18 '21 at 11:54