If all the roots of a polynomial P(z) have negative real parts, prove that all the roots of P'(z) also have negative real parts

Question

If all the roots of a polynomial $P(z)$ have negative real parts, prove that all the roots of the derivative $P'(z)$ also have negative real parts.

Could anyone provide a proof for this please?

Answer 1

Let $P(z)=\prod_{i=1}^n(z-\alpha _i)$ with $\Re \alpha _i<0$ $(i=1,2,...,n)$ and $$g(z)=\frac{1}{z-\alpha_1}+\frac{1}{z-\alpha _2}+...+\frac{1}{z-\alpha _n}.$$ Then $P^\prime (z)=P(z)g(z)$, and hence every root $z_0$ of $P^\prime(z)=0$ satisfies $P(z_0)=0$ or $g(z_0)=0$.
Suppose that $g(z_0)=0$ and $\Re z_0\ge 0$. It is easy to see that $\Re \frac{1}{z_0-\alpha _i}>0 $ $(i=1,2,...,n)$. Thus we have $$\Re g(z_0)=\Re\frac{1}{z_0-\alpha _1}+\Re\frac{1}{z_0-\alpha _2}+...+\Re\frac{1}{z_0-\alpha _n}>0,$$ which contradicts that $g(z_0)=0$.
Therefore all the roots of $ P^\prime (z)$ have negative real parts.