# One Question .Complex Numbers

1. Jul 7, 2008

### selena

One Question.....Complex Numbers

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

verify that: Square root (2) * lzl >= l Re (z) l + l Im ( z ) l

suggestion : reduce this inequality to ( lxl - lyl )^2 >=0

note : lxl <<<< modulus x

2. Jul 7, 2008

### HallsofIvy

Staff Emeritus
Re: One Question.....Complex Numbers

Well, what have you TRIED? If z= x+ iy, what is |z|? Whar are Re(z) and Im(z)?

I assume your note simply means that |x| and "modulus x" are the same thing. I had first interpreted it as "much less than" which makes no sense!

3. Jul 7, 2008

### selena

Re: One Question.....Complex Numbers

lxl <<<< modulus x

<<<< i dosent mean less thab but i mean modulus x ---->lxl
i mean modulus x = lxl

((2x2+2y2)1/2 - lxl - lyl )2 >= 0

Substitute lxl - lyl = k
2x2 + 2y2 - 2k * (2x2 + 2y2)1/2 + k2 >= 0

Substitute 2x2 + 2y2 = m

m - 2k * m1/2 + k2 >= 0

this is my try ...
i know its wrong ...
and i cant continue ...

Last edited: Jul 7, 2008
4. Jul 7, 2008

### Dick

Re: One Question.....Complex Numbers

You're going the wrong way. From your attempt it looks like you know |z|=sqrt(x^2+y^2) for z=x+iy. Start with what you want to prove. sqrt(2)*sqrt(x^2+y^2)>=|x|+|y|. Square both sides. Now move everything to one side and look at your suggestion.

5. Jul 7, 2008

### tiny-tim

Welcome to PF!

Hi selena! Welcome to PF!

(have a square root: √ and a square: ² )

Do you mean 2(x² + y²) ≥ (|x| + |y|)²?

hmm … let's keep it simple …

Hint: 2(x² + y²) = |x|² + |x|² + |y|² + |y|²