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

Maybe start with the left side

| |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?

 

[micromass]

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.

 

[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.

 

[Jamin2112]

 

[micromass]

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.

 

[micromass]

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 a_j = x_j + iy_j, b_j = u_j + iv_j

and after some simplification came up with

∑(x_j²+b_j²)(u_j²+v_j²) = ∑(x_j²+b_j²)∑(u_j²+v_j²)

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