# Complex numbers moduli sum

cheeez
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

Are there any restrictions on c? As I interpret the statement, c = 1/2 will always work.

cheeez
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.

Homework Helper
Gold Member
You can obtain 1/2 by considering n=2. For any n, you can gather the n terms together into 2 terms.

cheeez
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.

Homework Helper
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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.

cheeez
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

Homework Helper
Gold Member
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

$$\sum_k |z_k| = \left( |z_1|+\cdots+|z_i| \right) + \left( |z_{i+1}|+\cdots+|z_n| \right).$$

cheeez
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

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Then use the triangle inequality to show that

$$|z_1+\cdots+z_i| + |z_{i+1}+\cdots+z_n| \geq 1.$$

cheeez
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

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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.

cheeez
yea that's what I'm trying to do, any ideas?

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Actually any partition will work. You have 2 positive numbers whose sum is $$\geq 1$$. There's only a couple of cases to consider. I can't really be anymore specific without completely giving the proof.

cheeez
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.

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Those don't satisfy $$|z_1+z_2 +z_3| \geq 1$$.

cheeez
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