I have to give a geometric argument that, given two complex numbers $z_1, z_2$, the following inequality holds $$|z_1-z_2|\ge ||z_1|-|z_2||$$

I know every complex number has a nonnegative modulus, and this becomes a problem if $|z_1|\lt |z_2|$, and it contradicts the fact that $|z|\ge 0$ for all $z \in \mathbb{C}$. Intuitively I understand what is happening, but I'm having trouble formulating a geometric argument. I would imagine that, looking at a Argand diagram, the Triangle Inequality would be $$|z_1|+|-z_2|\ge |z_1-z_2|$$ which makes sense. But I'm not sure where to go from here.

  • $\begingroup$ I do not yet understand where the problem/contradiction should be. $\endgroup$
    – Martin R
    Sep 6 '15 at 18:37

If you want to proof it using the triangle inequality:

Apply the triangle inequality to get $|z_1-z_2|+|z_2| \geq |z_1|$ if $z_2\geq z_1$.
Apply it again to get $|z_2-z_1|+|z_1| \geq |z_2|$ if $z_1\geq z_2$.

This gives together with $|x|=|-x|$ and $|x|=x$ if $x \in \mathbb R_{\geq 0}$ the desired inequality.

  • $\begingroup$ I'm looking for a geometric approach. Would the Triangle inequality be a geometric approach or algebraic? My gut says algebraic, so how would I reformulate to prove geometrically? $\endgroup$
    – Lalaloopsy
    Sep 6 '15 at 18:47
  • 2
    $\begingroup$ There is an algebraic version of the triangle inequality and a geometric version. It is not that hard to switch form the one to the other. Try a triangle in the complex plane with coordinates $(0,0)$, $(\Re(z_1),\Im(z_1))$, $(\Re(z_2),\Im(z_2))$. Analytical geometry is geometry. $\endgroup$
    – wythagoras
    Sep 6 '15 at 18:56

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