Is it possible to reduce this inequality to (|x| - |y|)² ≥ 0?

[selena]

One Question...Complex Numbers

Homework Statement

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


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

note : lxl <<<< modulus x

 

[HallsofIvy]

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!

 

[selena]

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 * m^1/2 + k² >= 0

this is my try ...
i know its wrong ...
and i can't continue ...

 

[Dick]

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.

 

[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