Complex number vector prove

Question

Prove that, where $z$ denotes complex numbers:

a.) $|z_1 - z_2| \le |z_1| +|z_2|$

b.) $||z_1| - |z_2|| \le |z_1 +z_2|$

For a. the book answer gives: $|z_1 -z_2| = |z_1 + (-z_2)| \le |z_1| +|-z_2| =|z_1| +|z_2|$, but that proof isn't helpful at all, to me, because it doesn't explain anything. How can I prove this using a clear explaination instead of the way the book did?

Answer 1

Using Mejrdad's hint, for example:

$$z_k:=x_k+y_ki\,\,\,,\,,k=1,2\,\,,\,x_k,y_k\in\Bbb R\Longrightarrow$$

$$ \begin{align*}(1)&\;\;\;\;\;\;|z_1-z_2|=|(x_1-x_2)+(y_1-y_2)i|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ (2)&\;\;\;\;\;\;|z_1|+|z_2|=\sqrt{x_1^2+y_1^2}+\sqrt{x_2^2+y_2^2}\end{align*}$$

So squaring both equations above:

$$x_1^2+y_1^2+x_2^2+y_2^2-2(x_1x_2+y_1y_2)\leq x_1^2+y_1^2+x_2^2+y_2^2+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\Longleftrightarrow$$

$$-x_1x_2-y_1y_2\leq\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}\stackrel{\text{squaring}}\Longleftrightarrow$$

$$ x_1^2x_2^2+y_1^2y_2^2+2x_1x_2y_1y_2\leq x_1^2x_2^2+y_1^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\Longleftrightarrow$$

$$(x_1y_2-x_2y_1)^2\geq 0$$

And since the last inequality is trivial we get what we want going backwards( Why is it possible to argue that way?)

Answer 2

I think you might be able to do it by letting z = a + bi and then using the triangle inequality to prove the given relationships.