Showing that $d(z,w)$ is a metric in $\mathbb C$

Question

Defining $d(z,w)$ as $d(z,w) = |z - w|$, where $z, w \in \mathbb C$. Show that $d(z,w)$ is a metric on $\mathbb C$.

I managed to prove the positive-definite quality and the symmetric property, but I am having trouble proving the triangle inequality: $$d(z, w) \leq d(z,u) + d(u,w),\quad\forall z,w,u \in \mathbb C$$ i.e.: $$|z-w| \leq |z-u| + |u-w|.$$ Any help or suggestions are greatly appreciated. Thanks!

Answer 1

It is sufficient to show $|z+w|\leq|z|+|w|$. This is equivalent to $|z+w|^2\leq(|z|+|w|)^2$. Let $z=x+iy$, $w=u+iv$. Then $$|z+w|^2=(x+u)^2+(y+v)^2=x^2+y^2+u^2+v^2 +2xu + 2yv$$ Now, $|xu+yv|^2\leq (x^2+y^2)(u^2+v^2)$ (Can you prove this?) and since $a\leq|a|$ always, we get $$|z+w|^2\leq (x^2+y^2)+(u^2+v^2)+2\sqrt{(x^2+y^2)(u^2+v^2)}=(|z|+|w|)^2$$

This was just a long way of sayig that the usual Euclidean distance in two dimensions satisfies the triangle inequality.

Answer 2

$|z-w|=|z-u+u-w|\leq |z-u|+|u-w|.$