Triangle inequality for complex numers

Click For Summary

Homework Help Overview

The discussion revolves around the triangle inequality for complex numbers, specifically the inequality |(|z|-|z'|)| <= |z-z'|. Participants are examining the validity of an attempted proof and exploring the implications of their algebraic manipulations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the expansion of complex numbers and the implications of squaring terms. There is a focus on identifying errors in algebraic steps and the importance of maintaining the correct sign during manipulation.

Discussion Status

The discussion is active, with participants seeking clarification on the original poster's algebraic approach. Some guidance has been offered regarding visualizing the problem geometrically using an Argand diagram, suggesting a connection to vector representation. The original poster has identified a mistake related to sign changes in their calculations.

Contextual Notes

There is an emphasis on the need for clear algebraic presentation to facilitate understanding among participants. The original poster's error involved multiplying by a negative number, which highlights the importance of careful manipulation in proofs involving inequalities.

pivoxa15
Messages
2,250
Reaction score
1

Homework Statement


show
|(|z|-|z'|)|<=|z-z'|




The Attempt at a Solution



I used z=a+ib and z'=a'+ib' and ended up with the reverse inequality to the above by proving (ab'-ba')^2>=0 hence the reverse of the sign above. Where have I gone wrong?
 
Physics news on Phys.org
Well, I don't know what you did wrong if I can't see what you did!
 
Good point. Bascially I expanded the complex numbers out and squared them.
 
Ok, you're not getting the point here. Show me exactly what you did, in algebra, and I can see whether you are correct or not.
 
Think of the numbers like vectors and draw them on an argand diagram...you'll get a triangle...and two sides added should be greater than the third side...so...vice versa
 
I found my mistake. I multiplied by a negative number along the way so the sign was flipped and I didn't account for that when I first did it.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K