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

Jamin2112

## 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?

Staff Emeritus
Homework Helper
Use that

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

Jamin2112
Use that

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

Stay around here. I'm going to hit you up with another question later.

Staff Emeritus
Homework Helper
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.

Jamin2112

Staff Emeritus
Homework Helper
This is better

Jamin2112
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.

Staff Emeritus
Homework Helper
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?

Jamin2112
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.