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    $\begingroup$ Maybe, the following related result of Kellog is helpful: Let $A$ be a complex $n\times n$ matrix. If all its elementary symmetric functions are positive (so that the characteristic polynomial has alternating signs), then the spectrum of $A$ lies in the set $\{z : |\text{arg}z| \le \pi - \pi/n\}$.... $\endgroup$
    – Suvrit
    Commented Aug 15, 2015 at 1:51
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    $\begingroup$ In the wedge case you can compute explicitly the coefficients and see they are positive. It's a particular case of Suffridge's extremal polynomials- you have the expression deduced in Sheil-Small, Complex polynomials, p. 251-252. $\endgroup$
    – user75485
    Commented Aug 18, 2015 at 20:42
  • $\begingroup$ In fact the Suffridge polynomials cover the more general case when $1\not\in S,$ S self-conjugate and the arguments of the roots in S are separated by the same angle, except for a pair. $\endgroup$
    – user75485
    Commented Aug 19, 2015 at 14:44
  • $\begingroup$ @Josep: The reference you gave is extremely interesting. I was totally unaware of Suffridge's result. $\endgroup$
    – Louis Deaett
    Commented Aug 21, 2015 at 20:02
  • $\begingroup$ "I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial": yes, you're correct. The easiest way to see this is plugging in x=1: it's easy to show that $\Phi_m(1) = \begin{cases}0 & m=1 \\ p & m=p^k \\ 1 & \text{otherwise}\end{cases}.$ This essentially says there must be cancellation outside of the prime-power case. $\endgroup$
    – Joshua P. Swanson
    Commented May 6, 2019 at 14:22