Let $P(z)$ be a polynomial with non-negative real coefficients. Suppose $P(z)$ has all its zeros in a sector $S $ with sector angle greater than $\pi$ contains all the zeros of $P(z).$ My intuition says that $P'(z)$ also has its all zeros in $S.$ Am I correct or wrong? If correct, how can I proceed with the proof ?
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Try $P(z) = z^2 + z + 1$. Its zeros are in the sector $-2\pi/3 \le \theta \le 2\pi/3$, but the zero of $P'$ is $-1/2$ which is not.
answered Nov 21, 2019 at 18:34
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$\begingroup$ Can we conclude that for non-convex sector, Lucas Theorem is not true? $\endgroup$ Commented Nov 25, 2019 at 7:09
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1$\begingroup$ For any non-convex open set $S$ there is a polynomial with its roots in $S$ whose derivative has a root not in $S$. Namely, take two points $a,b \in S$ such that $(a+b)/2 \notin S$, and the polynomial is $(z-a)(z-b)$. $\endgroup$ Commented Nov 25, 2019 at 16:02