Ahlfors gives the following proof for the Gauss-Lucas theorem:
Theorem: If all zeros of a polynomial $P(z)$ lie in a half plane, then all zeros of the derivative $P'(z)$ lie in the same half plane
From $P(z) = a_n(z – \alpha_1)(z – \alpha_2)\cdots(z – \alpha_n)$ we obtain $\frac{P'(z)}{P(z)} = \sum_{i=1}^n\frac{1}{z – \alpha_n}$. Suppose that the half plane $H$ is defined as the part of the plane where $\mathrm{Im}((z – a)/b) < 0$. If $\alpha_k$ is in $H$ and $z$ is not, we have then that $\mathrm{Im}((z – \alpha_k)/b) = \mathrm{Im}((z – a)/b) – \mathrm{Im}((\alpha_k – a)/b) > 0$. But the imaginary parts of the reciprocal numbers have opposite sign. Therefore, under the same assumption, $\mathrm{Im}(b(z – \alpha_k)^{-1}) < 0$. If this is true for all $k$, we conclude that $\mathrm{Im}(\frac{bP'(z)}{P(z)}) = \sum_{i=1}^n\mathrm{Im}(b(z – \alpha_k)^{-1}) < 0$ and consequently $P'(z) \neq 0$.
What I don't understand from this proof is that at what point do we show that the zeros of $P'(z)$ lie in the same half plane? The end statement shows that $P'(z)$ is non-zero in $H$, but that is not what we ought to show.
Best Answer
It proved the proposition by contrapositive. Ahlfors considered any arbitrary $z\notin H$, and showed that $\alpha_k\in H$ for all $k$ implies $P’(z)\neq0$ because in particular the imaginary part isn’t zero. So, all the zeroes $z$ of $P’$, if they exist, cannot have $z\notin H$ (therefore $z\in H$...). You maybe missed: “and $z$ is not”.