# Equality of triangle inequality in complex numbers

$z$ 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!

#### Solutions Collecting From Web of "Equality of triangle inequality in complex numbers"

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.