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.



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


