# Prove ||x|-|y||≤|x-y|, where x and y are complex

## Homework Statement

If x, y are complex, prove that

| |x| - |y| | ≤ |x - y|

## Homework Equations

If x = a + ib, |x| = √(a2+b2)

|x + y| ≤ |x| + |y| (works for both complex and real numbers)

## The Attempt at a Solution

| |x| - |y| | = | |x| + (-|y|) ||x| + |-|y|| = |x| + |y|

....... Maybe almost there is I can show |x| + |y| ≤ |x - y| ........

How can I not get this problem? Use that

$$|x|=|(x-y)+y|$$

Use that

$$|x|=|(x-y)+y|$$  Stay around here. I'm gonna hit you up with another question later.

What you wrote down is incorrect. Specifically, the first inequality is wrong.

You need to prove two things:

$$|x|-|y|\leq |x-y|~\text{and}~|x|-|y|\geq -|x-y|$$

These two together would imply your inequality. This is better This is better Think you could help me with the Schwartz equality problem? My homework is due in 1 hour and that's the only one I have left.

Think you could help me with the Schwartz equality problem? My homework is due in 1 hour and that's the only one I have left.

Well, what is the problem and what did you try?

Well, what is the problem and what did you try?

Figuring out under what condition equality holds in the Schwartz inequality. (I know the answer is when a and b are linearly independent)

I let aj = xj + iyj, bj = uj + ivj

and after some simplification came up with

∑(xj2+bj2)(uj2+vj2) = ∑(xj2+bj2)∑(uj2+vj2)

which somehow shows that a is a scalar multiple of b. Not sure how, though.