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

1. Oct 11, 2011

### Jamin2112

1. The problem statement, all variables and given/known data

If x, y are complex, prove that

| |x| - |y| | ≤ |x - y|

2. Relevant equations

If x = a + ib, |x| = √(a2+b2)

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

3. 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?

2. Oct 11, 2011

### micromass

Use that

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

3. Oct 12, 2011

### Jamin2112

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

4. Oct 12, 2011

### micromass

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.

5. Oct 12, 2011

### Jamin2112

6. Oct 12, 2011

### micromass

This is better

7. Oct 12, 2011

### Jamin2112

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.

8. Oct 12, 2011

### micromass

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

9. Oct 12, 2011

### Jamin2112

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.