Complex numbers moduli sum

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Main Question or Discussion Point

show there is a some constant c, independent of n, s.t. if {Z_j} are complex numbers and sum of |Z_j| from 1 to n >= 1, then there's a subcollection {Z_j_k} of {Z_j} s.t. sum of |Z_j_k| >= c.

Any hint on how to start this?
 

Answers and Replies

  • #2
mathman
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Are there any restrictions on c? As I interpret the statement, c = 1/2 will always work.
 
  • #3
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how did you get 1/2. No restrictions on C. I don't get the question fully this C is a number that works for any sequence of {Z_j} right? otherwise you can pick any subcollection calculate the modulus of the sum and pick some C less than that.
 
  • #4
fzero
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You can obtain 1/2 by considering n=2. For any n, you can gather the n terms together into 2 terms.
 
  • #5
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ok it's 1/2 because you are looking at the subcollection of only 1 term right. so at least 1 of 2 must exceed 1/2.

i dont see how you can divide them into 2 numbers because you would need to have |z_1+..+z_i| + |z_i+1+...+z_n| >= 1 but that's not necessarily true since triangle inequality works the other way.
 
  • #6
fzero
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There's no need to involve the triangle inequality. The theorem you stated in the OP is about the sum of moduli, not moduli of the sums.
 
  • #7
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i dont get it how are you dividing z_1,...,z_n into 2 terms so |z_1+..+z_i| + |z_i+1+...+z_n| >= 1
 
  • #8
fzero
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i dont get it how are you dividing z_1,...,z_n into 2 terms so |z_1+..+z_i| + |z_i+1+...+z_n| >= 1
Try

[tex] \sum_k |z_k| = \left( |z_1|+\cdots+|z_i| \right) + \left( |z_{i+1}|+\cdots+|z_n| \right). [/tex]
 
  • #9
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oh man I screwed up the original posting, I meant you need it so that |sum of z_j| >= C for some subcollection z_j of the original z_n
 
  • #10
fzero
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Then use the triangle inequality to show that

[tex] |z_1+\cdots+z_i| + |z_{i+1}+\cdots+z_n| \geq 1.
[/tex]
 
  • #11
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you're saying this is always true for any i? if you have 3 numbers each of which has length 1/3 and if you divide it into z1+z2 and z3 where z1 and z2 point in the opposite directions this wouldnt be true. I dont think its as simple as the triangle inequality
 
  • #12
fzero
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No it's not true for any i. You have to show that there is some partition that satisfies the premise of the theorem.
 
  • #13
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yea that's what I'm trying to do, any ideas?
 
  • #14
fzero
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Actually any partition will work. You have 2 positive numbers whose sum is [tex]\geq 1[/tex]. There's only a couple of cases to consider. I can't really be anymore specific without completely giving the proof.
 
  • #15
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but obviously the {z1,z2} , {z3} partition does not work for n=3 take z1 , z2 to be a+bi, -a-bi, and z3 to be some other vector perpendicular to them and each length 1/3.
 
  • #16
fzero
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Those don't satisfy [tex]|z_1+z_2 +z_3| \geq 1[/tex].
 
  • #17
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I think I see the confusion, my correction only applies to the 2nd part. the original problem should be

show there is a some constant c, independent of n, s.t. if {Z_j} are complex numbers and sum of |Z_j| from 1 to n >= 1, then there's a subcollection {Z_j_k} of {Z_j} s.t. |sum of Z_j_k| >= c.

so here |z1| + |z2| + |z3| = 1/3 + 1/3 + 1/3 = 1
 

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