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 ? 



Lucas Theorem on a sector
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.


