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Gauss-Lucas Theorem

 Theorem: If all zeros of a polynomial $P(z)$ lie in a half-plane, then all zeros pf the derivative $P'(z)$ lie in the same half-plane.

Proof: Let $P(z)$ be any polynomial with degree $n$. Then $$P(z)=a(z-a_1)(z-a_2)\cdots(z-a_n)$$ where $a_1,a_2,\dots,a_n$ are the zeros of $P(z)$. Hence $$\frac{P'(z)}{P(z)}=\frac1{z-a_1}+\frac1{z-a_2}+\cdots+\frac1{z-a_n}$$Suppose the half plane $H$ defined as the part of the plane where $$IM\frac{z-a}{b}<0$$. Suppose $z$ If $a_k$ is in $H$ and $z$ is no, we have then $$Im\frac{z-a_k}{b}=Im\frac{z-a}{b}-Im\frac{a_k-a}{b}>0$$But the imaginary parts of reciprocal numbers have opposite signs. Therefore, under the same assumption, $Im\ b(z-a_k)^{-1}<0$.Now  this is true for all $k$ we conclude that $$Im\frac{bP'(z)}{P(z)}=\sum_{k=1}^{n} Im\frac{b}{z-a_k}<0$$ and consequently $P'(z)\neq 0$. Hence $z$ is not a root of $P'(z)$ concluding that all roots of $P'(z)$ lie in $H$. $\blacksquare$

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