# Convex polygon containing the zeros of a convex linear combination of polynomials

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.

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.