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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?

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  • $\begingroup$ 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. $\endgroup$ – fleablood Jun 23 '16 at 0:17
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There's two things about your question that need to be addressed.

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 $\overline{z} = z$.

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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

Complex addition

w + z = (a + x) + (b + y)i

Complex absolute value

|w| = (a^2 + b^2)^1/2

Write out the formulas for |w| + |z|, and |w + z| in terms of a, b, x, and y. Then apply your squaring technique, and you'll have it.

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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?

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