How to prove that all zeros of the complex polynomial $P(z)$ lie in $\Bbb{D}$?
I want to know how to prove that all zeros of the polynomial $P(z)$ lie in the closed unit disk $|z| \leqslant 1$. Where $$P(z)=z^{n+1}+\frac{2(n+1)\cos\alpha}{n+2}z^{n}+\frac{n}{n+2}z^{n-1}+\frac{2 e^{-i \theta}}{n+2}z +\frac{2e^{-i \theta}\cos\alpha}{ n+2}$$ $n\geqslant 1,\pi/2 \leqslant\alpha < \pi,\theta\in \mathbb{R}$.
We can write $P(z)$ as
$$\begin{split}
P(z)&=\left(z^{n+1}+\frac{n}{n+2}z^{n-1}+\frac{2 e^{-i \theta}}{n+2}z\right) +\left(\frac{2(n+1)\cos\alpha}{n+2}z^{n}+\frac{2e^{-i \theta}\cos\alpha}{ n+2}\right)\\
&=:\psi(z)+\phi(z)
\end{split}$$
where $\psi(z)=z^{n+1}+\frac{n}{n+2}z^{n-1}+\frac{2e^{-i\theta}}{n+2}z$ and $\phi(z)=\frac{2(n+1)\cos\alpha}{n+2}z^{n}+\frac{2e^{-i\theta}\cos\alpha}{n+2}$.
By Cohn's Rule, we can prove that all zeros of $\psi(z)$ are lie in the closed unit disk $|z| \leqslant 1$. On the unit circle $|z|=1$, since $\frac{\pi}{2} \leqslant \alpha<\pi$, one can find that
\begin{split}
|\psi(z)|=\left|\frac{2(n+1+e^{-i\theta})}{n+2}\right|>\left|\frac{2(n+1+e^{-i\theta})\cos\alpha}{n+2}\right|=|\phi(z)|.
\end{split}
Then by Rouche’s theorem, we know that $\psi(z)$ and $P(z)=\psi(z)+\phi(z)$ have the same number of zeros inside the unit circle $|z|=1$.
Now my amazement is if some zeros of $\psi(z)$ lie on the unit cicle, can we show that some zeros of $P(z)$ lie on the unit cicle $|z|=1$, also? In other words, If all zeros of $\psi(z)$ lie in the closed unit disk $|z|\leqslant 1$, can we prove that all zeros of $P(z)$ lie in the closed unit disk $|z|\leqslant 1$?
For example: If we take $n=10, \cos\alpha=-\frac{5}{6},e^{-i\theta}=i$, then $\psi(z)=z^{11}+\frac{5}{6}z^{9}+\frac{i}{6}z$ and $P(z)=z^{11} - \frac{55}{36} z^{10} + \frac{5}{6} z^9 + \frac{i}{6}z - \frac{5i}{36}$ , and all zeros distribution of $\psi(z)$ and $P(z)$ are illustrated by the following images.
Zeros distribution of
$\psi(z)$
Zeros distribution of
$P(z)$
We can see that all zeros of $\psi(z)$ and $P(z)$ lie in the closed unit disk $|z|\leqslant 1$. But how to prove it? Ask for help!
Notice that $P(z) = \frac{1}{n+2} Q'(z)$, where $Q(z) = (z^n + e^{-i\theta})(z^2 + 2z\cos\alpha + 1)$. The roots of $(z^n + e^{-i\theta})$ then obviously lie on the unit circle, and the roots of $(z^2 + 2z\cos\alpha + 1)$ are given by $r = -\cos\alpha \pm i\sin\alpha$ via the quadratic formula. So, those roots lie on the unit circle as well. With this, the roots of $Q(z)$ are all on the unit circle, and by Gauss-Lucas the roots of $P(z)$ are all in the closed unit disc.
Finally, this also tells you where $P(z)$ has roots on the unit circle: it's precisely where $(z^n + e^{-i\theta})$ and $(z^2 + 2z\cos\alpha + 1)$ share a root.