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

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Homework Help Overview

The discussion revolves around verifying an inequality involving complex numbers, specifically relating the modulus of a complex number to its real and imaginary parts. The original poster suggests reducing the inequality to a form involving squared differences of absolute values.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the modulus of a complex number and its components, questioning the interpretation of terms like "modulus" and "absolute value." There are attempts to manipulate the inequality algebraically, with some participants suggesting starting points and substitutions.

Discussion Status

The discussion is active, with various attempts to approach the problem and clarify terms. Some participants have provided hints and suggestions for reformulating the inequality, while others express uncertainty about their attempts and seek further guidance.

Contextual Notes

There is a noted confusion regarding the notation and terminology used, particularly around the concept of modulus and its relation to the components of complex numbers. Participants are also navigating the constraints of the homework problem without providing direct solutions.

selena
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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
 
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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!
 


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 can't continue ...
 
Last edited:


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.
 
Welcome to PF!

Hi selena! Welcome to PF! :smile:

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

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

hmm … let's keep it simple … :rolleyes:

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

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