# Complex number vector prove

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?

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## 2 Answers

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?)

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1
+1 It looks fine and nice. I don't understand why some tutors want the students not to use some well-known facts when they solve the routine problems. In fact, the triangle inequality is a very basic tool when we face these kinds of inequalities. Any way, Nice illustration, Don. – Babak S. Jan 18 at 4:01
Thanks a lot Don! – Q.matin Jan 18 at 6:20

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.

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1
But, the OP, was suggested to use the same method. Do you think your approach would change the way? – Babak S. Jan 18 at 3:22
@BabakSorouh: It's the same thing, I just explained that it's the triangle inequality since the book seemed to have given no explanation of why it's true. – Mehrdad Jan 18 at 3:24
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