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Let $p(z)=\prod_{k=1}^n(z-z_k)$ and $p_k(z)=\prod_{i=1,i\neq k}^n(z-z_i). $ Then $p'(z)=\sum_{k=1}^np_k(z).$ Let $q(z)=(1/n)p'(z)= (1/n)\sum_{k=1}^np_k(z).$ Suppose $p(z)$ has all its zeros in a convex polygon $C.$ Then by Gauss-Lucas Theorem $q(z)$ has all its zeros in $C.$ Now $q(z)$ can be thought of as a particular case of convex linear combination of $p_k, \;1\leq k\leq n,$ i.e., $q(z)=Q(z)$ if we take each $a_k=1/n$ in $Q(z)=\sum_{k=1}^na_kp_k(z),$ where each $a_k$ in $Q(z)$ in general is a non-negative real such that $\sum_{k=1}^na_k=1.$ I think $Q(z)$ also has all its zeros in $C.$ Can the proof of Gauss Lucas Theorem be used to prove this? Kindly suggest.

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Yes it can. If $z$ is a root of $Q(z)$ outside $C$, then $\sum a_k/(z-z_k)=0$, therefore (take the complex conjugate) we get $\sum a_k(z-z_k)/|z-z_k|^2=0$, but all summands in LHS belong to the same half-plane.

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  • $\begingroup$ Can Q(z) be further generalized? $\endgroup$ – user159888 Dec 10 '19 at 0:44

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