Proof of Jensen’s theorem on roots and circles

calculuspolynomialsroots

Show that if a polynomial $P(z)$ is a real polynomial not identically constant, then all nonreal zeros of $P'(z)$ lie inside the Jensen disks determined by all pairs of conjugate nonreal zeros of $P(z)$.

I found some sources that call it "Jensen's theorem".

I tried to prove it using logarithmic derivative and finding a contradiction but I don't find it.

Could you please provide me some hints (without use of Gauss Lucas theorem) ?

Best Answer

If $z$ lies outside the closed Jensen disc between $\alpha$ and $\overline{\alpha}$ then the angle between $z-\alpha$ and $z-\overline{\alpha}$ is acute. That is, $$\operatorname{Re}\left((z-\alpha)(\overline z - \alpha)\right)> 0.$$ If you also take $\operatorname{Im}(z) > 0$ then $$\begin{eqnarray} \operatorname{Im}\left(\frac1{z-\alpha} + \frac1{z-\overline{\alpha}}\right) &=& \operatorname{Im}\left(\frac1{z-\alpha} - \frac1{\overline z-\alpha}\right) \\[1ex] &=& \operatorname{Im} \left( \frac{\overline z - z}{(z - \alpha)(\overline z - \alpha)} \right) \\[1ex] &=& \operatorname{Im} \left( \frac{-2 \mathrm i \operatorname{Im}(z)}{(z - \alpha)(\overline z - \alpha)} \right) \\[1ex] &=& -2 \operatorname{Im}(z) \operatorname{Re} \left( \frac1{(z - \alpha)(\overline z - \alpha)} \right) \\[1ex] &<& 0. \end{eqnarray}$$ (Where we use the fact that $\operatorname{Re}(w) > 0$ if and only if $\operatorname{Re}(w^{-1}) > 0$.) Now apply this to the logarithmic derivative of $P$.

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