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I'm trying to prove that if $z,w$ are non non-zero complex numbers then $|z+w|=|z|+|w|$ iff there exists a positive real $c$ st $z=cw$.

Now I've proved the leftward implication but I'm having difficulties with the rightward one, so I'd appreciate any hint.

NOTE: I can't use the exponential or polar representation of complex numbers nor the notion of scalar product but only the basic algebraic properties of complex numbers.

lorenzo
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  • $|cw| + |w| = c|w| + |w| = (c+ 1)|w| = |(c+1)w| = |cw + w| = |z +w|$. –  Nov 09 '17 at 10:27
  • @A---B I already proved this. – lorenzo Nov 09 '17 at 10:30
  • Related: https://math.stackexchange.com/q/2011729/42969. – Martin R Nov 09 '17 at 10:35
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    @lorenzo Assume $z = c+ id, w = a+ib$. $\sqrt{(a+c)^2 + (b + d)^2} = \sqrt{a^2 + b^2} + \sqrt{d^2 + c^2} \iff (a+c)^2 + (b + d)^2 = a^2 + b^2 + d^2 + c^2 + 2\sqrt{a^2 + b^2}\sqrt{d^2 + c^2}\ \iff (ac + bd)^2 = (a^2 + b^2)(c^2 + d^2) \iff 2abcd = (ad)^2 + (bc)^2 \iff 0 = (ad - bc)^2 \iff bc = ad \iff c = \frac{ad}{b} $, therefore $ z = d ((a/b) + i) = (d/b)(a + ib) = (d/b)w = cw$. –  Nov 09 '17 at 13:12
  • counterexample: $z=1$, $w=0$. – David C. Ullrich Jul 18 '21 at 11:27

2 Answers2

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If $|z+w|=|z|+|w|$, then their squares are equal. But\begin{align}|z+w|^2=\bigl(|z|+|w|\bigr)^2&\iff(z+w)\bigl(\overline z+\overline w\bigr)=z.\overline z+w.\overline w+2|z|.|w|\\&\iff z.\overline z+z.\overline w+w.\overline z+w.\overline w=z.\overline z+w.\overline w+2|z|.|w|\\&\iff z.\overline w+w.\overline z=2|z|.|w|\\&\iff\operatorname{Re}\bigl(z.\overline w\bigr)=|z|.|w|.\end{align}Now, if you write $z$ and $w$ as $a+bi$ and $c+di$ respectively (with $a,b,c,d\in\mathbb R$), the previous equality says that $(a,b).(c,d)=\bigl\|(a,b)\bigr\|.\bigl\|(c,d)\bigr\|$. Therefore, if $\theta$ is the angle between $c$ and $d$, then $\cos\theta=1$ and so $z=cw$, for some $c\in[0,+\infty)$.

José Carlos Santos
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Let $z=r(\cos t+i\sin t), w=s(\cos u+i\sin u)$ where $r(\ge0),s(\ge0),t,u$ are real

$|z+w|^2=r^2+s^2+2rs\cos(t-u)$

Now $(|z|+|w|)^2=r^2+s^2+2rs$

So, we need $\cos(t-u)=1\iff t\equiv u\pmod{2\pi}$

lab bhattacharjee
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  • I can't use the exponential or polar representation of complex numbers nor the notion of scalar product – lorenzo Nov 09 '17 at 13:10