# Triangle inequality for complex numbers

I just start to learn about complex numbers and I want to prove the triangle inequality, which says that if $z$ and $w$ are complex numbers, then $\displaystyle |z + w| \le |z| + |w|.$ My approach is to square both sides of the inequality (since each side is nonnegative) to obtain the equivalence $|x| \ge x$ for every $x \in \mathbb{C}.$ Now squaring the right hand side yields $|z|^2 + 2|z||w| + |w|^2,$ but for the left hand side, why doesn't it hold that $|z + w|^2 = (z + w)^2 = z^2 + 2zw + w^2$ like with real numbers?

• In real numbers if |x|=b there are only two possible values for x. Either b or -b. In both cases x^2 = b^2. In complex numbers there are an infinite number of solutions to |x| =b. But only two of them have x^2 = b^2. Example. |x|=1 has solutions, 1,-1,i,-i, 1/ sqrt2 + i/sqrt 2 etc. But only 2 of them are actually roots of 1. So in general it simply isn't true that |x|^2 =x^2. – fleablood Jun 23 '16 at 0:17

It might be easier if you said $z = x + i y, w = a + bi$

$\sqrt{x^2 + y^2} + \sqrt{a^2 + b^2} \ge \sqrt {(x+a)^2 + (y+b)^2}$ And now you are in real numbers, square both sides.

otherwise:

$|z|^2 = z\bar z\\ \overline{(z + w)} = \bar z + \bar w\\ |z+w|^2 = (z+w)(\bar z + \bar w) = |z|^2 + z\bar w + \bar z w + |w|^2$

And: $z\bar w + \bar z w = 2Re (zw)$

But would I be able do show you that without breaking out $z$ and $w$ as in the first example?

1) $|x| \geq x$ doesn't quite match your intuition for what it means, since $x \geq y$ comparisons require you to be in $\mathbb{R}$. For example, is $j \geq 1$? (You can define partial orders and such, but I don't think that's what you meant.)

2) $|z + w|^2 = (z + w) (\overline{z + w})$, where $\overline{z}$ is the complex conjugate of $z$. This is a generalization of what it means in the real case, where $\overline{z} = z$.

You're on the right track, just start with the definitions and convert the problem to Real numbers

Given:

w = a + bi

z = x + yi