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Math Help - Location of complex roots of z^n + a_n-1 z^n-1 + ... + a_0 where all a_k in [0,1]

  1. #1
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    Location of complex roots of z^n + a_n-1 z^n-1 + ... + a_0 where all a_k in [0,1]

    Problem:
    Let P: \mathbb{C} \rightarrow \mathbb{C}, \hspace P(z)=z^n + \sum_{k=0}^{n-1}a_{k}z^{k} be a polynomial of degree n\in\mathbb{N} with real coefficients a_k \in [0,1] for k=0,\dots,n-1.

    Show that if P(z_0)=0 for some z_0 \in \mathbb{C} then Re(z_0)<0 or |z_0|<(1+\sqrt{5})/2.

    Ideas :

    I'm kind of stuck on this one.

    I observed that there are clearly no roots on the positive real axis. I could use Descarte's Rule of Signs if necessary.

    I also know, via the same method described here, that all roots have modulus less than or equal to 2.

    I was thinking of writing z_0=r e^{i\theta} and using the statement that Re(P(z_0))=0=Im(P(z_0)) to get some relationship between r and \theta using trig functions but it hasn't panned out for me so far.

    I also wonder if Vieta's formula for the coefficients in terms of the roots might help.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Location of complex roots of z^n + a_n-1 z^n-1 + ... + a_0 where all a_k in [0,1

    Quote Originally Posted by robertoCamerani View Post
    Problem:
    Let P: \mathbb{C} \rightarrow \mathbb{C}, \hspace P(z)=z^n + \sum_{k=0}^{n-1}a_{k}z^{k} be a polynomial of degree n\in\mathbb{N} with real coefficients a_k \in [0,1] for k=0,\dots,n-1.

    Show that if P(z_0)=0 for some z_0 \in \mathbb{C} then Re(z_0)<0 or |z_0|<(1+\sqrt{5})/2.

    Ideas :

    I'm kind of stuck on this one.

    I observed that there are clearly no roots on the positive real axis. I could use Descarte's Rule of Signs if necessary.

    I also know, via the same method described here, that all roots have modulus less than or equal to 2.

    I was thinking of writing z_0=r e^{i\theta} and using the statement that Re(P(z_0))=0=Im(P(z_0)) to get some relationship between r and \theta using trig functions but it hasn't panned out for me so far.

    I also wonder if Vieta's formula for the coefficients in terms of the roots might help.
    That is not what we can define as 'elementary problem'. Odlyzko and Poonen proved some interesting things about the set of all roots of all polynomials with coefficients 0 or 1, and in particular they proved that the set of zeros of these polynomials is contained in the half-plane \text{Re}\ (z) < \frac{3}{2} and in the annulus  \frac{1}{\phi}<|z|< \phi where \phi= \frac{1+\sqrt{5}}{2} is the 'golden ratio'. The original paper can be found here...

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    Kind regards

    \chi \sigma
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    Re: Location of complex roots of z^n + a_n-1 z^n-1 + ... + a_0 where all a_k in [0,1

    Thanks for the link and the background. I had no idea this problem was referencing a much broader topic. I'll take a look over this paper.

    As for answering my question as stated, I came up with the following :

    It suffices to show that Re(z_0)>=0 \Rightarrow |z_0|<(1+\sqrt{5})/2.
    We note that |0|<(1+\sqrt{5})/2 so we can forget about that case from now on (and divide by z_0 if we need to).
    Write z_0 as z_0=re^{i\theta}
    So 0=P(z_0)=r^n e^{in\theta} + \sum_{k=0}^{n-1}{a_k r^k e^{ik\theta}.
    We multiply both sides by e^{-in\theta} to get
    0=r^n  + \sum_{k=0}^{n-1}{a_k r^k e^{i(k-n)\theta}.
    We then take the real part and extract the first term from the summation to get
    0=r^n  + a_{n-1} r^{n-1}\cos(-\theta)+\sum_{k=0}^{n-2}{a_k r^k \cos((k-n)\theta)}.
    Then, since 0<Re(z_0)=r\cos(\theta)=r\cos(-\theta) we know \cos(-\theta)\ge0.
    We then take the following steps
    -(r^n  + a_{n-1} r^{n-1}\cos(-\theta))=\sum_{k=0}^{n-2}{a_k r^k \cos((k-n)\theta)}
    r^n\le|(r^n  + a_{n-1} r^{n-1}\cos(-\theta)|=|\sum_{k=0}^{n-2}{a_k r^k \cos((k-n)\theta)}|<\sum_{k=0}^{n-2}r^k=\frac{1-r^{n-1}}{1-r}
    (The last inequality is strict because equality would only be attained if \forall k=0,\dots,n-2 a_k=1 \text{ and } \cos((k-n)\theta))=1 which implies \theta=0, but this polynomial has no positive real roots by Descarte's Rule of Signs.)

    Proceeding from the strict inequality we get
    if r\le1 we are fine, otherwise 1<r and so multiplying both sides by (1-r) not changes the direction of the inequality, giving
    r^n-r^{n+1}>1-r^{n-1}.
    Rearrange as
    0>-1<r^{n+1}-r^n-r^{n-1}.
    Divide both sides by r^{n-1} to get 0>r^2-r-1 which has roots (1\pm\sqrt{5})/2 and so |z_0|<(1+\sqrt{5})/2.
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