Prove that if a polynomial $P$ has no roots in the upper half plane, then so does $P'$

Question

Prove that if a polynomial $P$ has no roots in the upper half plane, then so does $P'$

This is a part of an exam preparation and I would appreciate a hint. My approach was to use Rouche's theorem but it didn't help so far.

Thanks!

Answer 1

Since the fact stated in the question is used to prove the Gauss-Lucas theorem, it should have a proof independent of that theorem. Which is easy to give: write $P(z)=c\prod_{k=1}^n (z-z_k) $, then $$\frac{P'(z)}{P(z)}=\sum_{k=1}^n \frac{1}{z-z_k}\tag1$$

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The above is a hint.

Complete proof is below.

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By assumption, $\operatorname{Im}z_k\le 0$ for all $k$. When $\operatorname{Im}z>0$, we have $\operatorname{Im}(z-z_k)>0$ and therefore $\operatorname{Im}\frac{1}{z-z_k}<0$ for all $k$. The conclusion is $$\operatorname{Im}\frac{P'(z)}{P(z)}<0\tag2$$

Answer 2

According to the Gauss-Lucas theorem, the zeroes of $P'$ are contained in the convex hull of the zeroes of $P$.