[complex functions] polynomial roots in a convex hull

AI Thread Summary
The discussion revolves around proving that the roots of the derivative of a polynomial W(z) with complex coefficients lie within the convex hull of its original roots. The poster, rahl, seeks clarification on the problem and confirms their understanding of the convex hull definition. They propose a solution involving Vieta's formulas and express concerns about the complexity of their calculations. Rahl indicates they have found the solution and plans to share it for the benefit of others. The thread highlights the connection between polynomial roots and their derivatives in complex analysis.
rahl___
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Hi everyone,

I've got this problem to solve:

Let W(z) be a polynomial with complex coefficients and complex roots. Show that the roots of W'(z) are in a convex hull of the set of roots of W(z).

My problem is that I don't fully understand the question.

I have found such definition of convex hull:
Given a set of points (z_1, z_2, ..., z_n), we denote convex hull as:
conv(z_1, z_2, ..., z_n) = \{ z = \sum_{k=1}^n \beta_k z_k : \beta_k \in [0,1], \sum_{k=1}^n \beta_k = 1 \}
So I do have to prove, that all the roots of W'(z) [let's denote them as z'_k] must be able to be written in such form:
z'_k = \sum_{k=1}^n \beta_k z_k, where \beta_k are satysfying the conditionsof convex hull and z_k are the roots of W(z).
Am i right?

If so, I thought about this kind of sollution:
we assume that what we have to prove is true, so we can write the roots of W'(z) as:
z'_j = \sum_{k=1}^n \beta_k^{(j)} z_k
Now we write down W'(z) using viete's formulas:
W'(z) = n a_n z^{n-1} - n a_n ( \sum_j \sum_k \beta_k^{(j)} z_k ) z^_{n-2} + n a_n \sum_{i<j} ( \sum_k \beta_k^{(i)} z_k ) ( \sum_k \beta_k^{(j)} z_k ) z^{n-3} - ... + n a_n \prod_j \sum_k \beta_k^{(j)} z_k [dont know why the second part of the equation landed little higher that the first one, sorry for that]
and compare it to the polynomial we get when multyplying W(z) by \sum_k {1} / (z-z_k). What do you think of it? I've tried to do this, but the calculus grow pretty vast and I feel that there is a simplier method of proving this.

I would appreciate if you could tell me wheter I understand the question right and if my idea of solving it looks fine.

thanks for your time,
rahl.
 
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I have found the solution. If no one deletes this thread, I will write how to solve it, maybe it will help someone some day.
 
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