Proving on modulus and conjugates

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Homework Help Overview

The discussion revolves around proving the inequality involving the modulus of a complex number and its real and imaginary components, specifically the assertion that \(\sqrt{2} |z| \geq |Re z| + |Im z|\). Participants are exploring the validity of certain inequalities and strategies in the context of complex analysis.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are questioning the validity of using the inequalities |Re z| ≤ |z| and |Im z| ≤ |z| to derive |Re z| + |Im z| ≤ 2|z|. There is a discussion about the implications of these inequalities and whether they can lead to the desired result.

Discussion Status

Some participants have suggested alternative approaches, such as squaring both sides of the inequality, and there is an acknowledgment that the initial strategy may not yield the least upper bound needed for the proof. The conversation is ongoing with different interpretations being explored.

Contextual Notes

There is a mention of needing to express z in terms of its components, specifically z = x + iy, which may be relevant to the proof being discussed.

irony of truth
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I am proving that (sqrt)2 |z| >= |Re z| + |Im z|. The professor gave this as an example, but I want to ask something...

Why is it that I can't use this strategy:
-> |Re z| <= |z|
-> |Im z| <= |z|
-> adding corresponding expression yields |Re z| + |Im z| <= 2|z|. (what's wrong here?)
 
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irony of truth said:
Why is it that I can't use this strategy:
-> |Re z| <= |z|
-> |Im z| <= |z|
-> adding corresponding expression yields |Re z| + |Im z| <= 2|z|. (what's wrong here?)

You can't use it because [itex]|Re z|+|Im z| \leq 2|z|[/itex] simply does not imply that [itex]|Re z| + |Im z| \leq \sqrt{2}|z|[/itex]. Simple example: Say [itex]|Re z|+|Im z|=1.5[/itex]. The first inequality is satisfied, but the second is not.
 
Try squaring both sides.

irony of truth said:
I am proving that (sqrt)2 |z| >= |Re z| + |Im z|. The professor gave this as an example, but I want to ask something...
Why is it that I can't use this strategy:
-> |Re z| <= |z|
-> |Im z| <= |z|
-> adding corresponding expression yields |Re z| + |Im z| <= 2|z|. (what's wrong here?)

That is an upper bound for |Re z| + |Im z|, but not the least upper bound. Try squaring both sides.
 
Oh yeah, put z=x+iy first.
 

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