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

  • Thread starter Jamin2112
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  • #1
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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?

feels-bad-man-.jpg
 

Answers and Replies

  • #2
22,129
3,297
Use that

[tex]|x|=|(x-y)+y|[/tex]
 
  • #3
986
9
Use that

[tex]|x|=|(x-y)+y|[/tex]

screen-capture-2-23.png



feelsgoodman.png










Stay around here. I'm gonna hit you up with another question later.
 
  • #4
22,129
3,297
What you wrote down is incorrect. Specifically, the first inequality is wrong.

You need to prove two things:

[tex]|x|-|y|\leq |x-y|~\text{and}~|x|-|y|\geq -|x-y|[/tex]

These two together would imply your inequality.
 
  • #6
22,129
3,297
This is better :smile:
 
  • #7
986
9
This is better :smile:

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
22,129
3,297
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?
 
  • #9
986
9
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.
 

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