# Homework Help: [complex functions] polynomial roots in a convex hull

1. Mar 25, 2009

### rahl___

Hi everyone,

I've got this problem to solve:

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

I have found such definition of convex hull:
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?

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.