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$z$ and $w$ be nonzero complex numbers. How do I show that $|z+w|=|z|+|w|$ if and only if $z=sw$ for some real positive number $s$.

I approached this by letting $z=a+ib$, and $w=c+id$, and kinda play around with it. I also tried to square both sides when proving forward direction, but I could not get it to work. Can anyone get me some ideas, maybe?

Thank you!

Martin Sleziak
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Akaichan
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2 Answers2

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Here's a cleaner proof of the second direction, which avoids coordinates.

Suppose $|z + w| = |z| + |w|$. Then $$ \begin{align*} |z|^2 + 2|zw| + |w|^2 &= (|z| + |w|)^2 \\ &= |z + w|^2 \\ &= (z + w)(\overline{z + w}) \\ &= z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w} \\ &= |z|^2 + 2\mathrm{Re}(z\overline{w}) + |w|^2. \end{align*} $$ This implies $|zw| = \mathrm{Re}(z\overline{w})$. But because $|zw| = |z\overline{w}|$, so $z\overline{w}$ must be a nonnegative real number. Hence, $\arg(z) = \arg(w)$, and we conclude $z = sw$ for some $s > 0$.

Gabriel Romon
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Frank
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First try to prove it in one direction.

Let $z = a + bi$ and $w = sa + sbi$. Then we have:

$$\mid z + w \mid = \mid a + bi + sa + sbi \mid = \mid |a(1+s) + (1+s)bi\mid = \sqrt{(1+s)^2a^2 + (1+s)^2b^2} = (1+s)\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2} + \sqrt{s^2a^2 + s^2b^2} = \mid z \mid + \mid w \mid$$

Now for the other direction. Let $z=a + bi$ and $w = c + di$. Then we have:

$$\mid z + w \mid = \mid a + bi + c + di \mid = \mid (a+c) + (b+d)i \mid = \sqrt{(a+c)^2 + (b+d)^2}$$

$$\mid z \mid + \mid w \mid = \mid a + bi \mid + \mid c + di \mid = \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}$$

Now since $\mid z + w \mid = \mid z \mid + \mid w \mid$ we have:

$$\sqrt{a^2 + c^2 + 2ac + b^2 + d^2 + 2bd} = \sqrt{a^2 + b^2} + \sqrt{c^2 + d^2}$$

Square both sides and we have:

$$a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = a^2 + b^2 + c^2 + d^2 + 2\sqrt{(a^2 + b^2)(c^2 + d^2)}$$

$$ac + bd = \sqrt{(a^2 + b^2)(c^2 + d^2)}$$

Now square again and multiply out:

$$a^2c^2 + b^2d^2 + 2abcd = a^2c^2 + b^2c^2 + a^2d^2 + b^2d^2$$ $$2abcd = b^2c^2 + a^2d^2$$ $$b^2c^2 + a^2d^2 - 2abcd = 0$$ $$(cb - ad)^2 = 0 \implies cb = ad$$

So let $c = sa$, then we have:

$$scb = sad \implies sb = d$$

Hence $z = a + bi$ and $w = c + di = as + sbi$. Q.E.D.

Stefan4024
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  • I think I got it! – Akaichan Jan 26 '14 at 19:30
  • @Akaichan I've posted a solution for the other direction too. Hope it helps. – Stefan4024 Jan 26 '14 at 19:32
  • Thanks! I got to the $cb=ad$ part, and was stuck for a bit. – Akaichan Jan 26 '14 at 19:55
  • Why does your $s$ near the end have to be positive? – Favst Feb 19 '21 at 15:49
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    @Favst If $s<0$ then we have that $ac <0$ and similarly $bd<0$. This is impossible as $\sqrt{(a^2+b^2)(c^2+d^2)} = ac+bd < 0$ is a contradiction. – Stefan4024 Feb 20 '21 at 00:49
  • Why is it legit to suppose $c=sa$? Shouldn't it be deduced somehow from $bc=ad$ without supposing anything? I feel like in the proof, $A\implies B$, you are starting with $A$ is true, but in the process, you suppose that part of $B$ is also true. It feels like cheating. – Mateo Jan 24 '24 at 12:02