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?
[Math] Triangle inequality for complex numbers
complex numbers
Best Answer
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?