Problem 1226

Let $f(z)=z^{n+1}-a z^{n}+a z-1$. If $\xi=\cos \phi+i \sin \phi$, we have
$$\begin{aligned} \xi^{n+1}-1 & =\cos (n+1) \phi-1+i \sin (n+1) \phi \\ & =-2 \sin ^{2} \frac{n+1}{2} \phi+2 i \sin \frac{n+1}{2} \phi \cos \frac{n+1}{2} \phi \\ & =-2 \sin \frac{n+1}{2} \phi \cdot\left(\sin \frac{n+1}{2} \phi-i \cos \frac{n+1}{2} \phi\right) \\ \xi^{n}-\xi & =(\cos n \phi-\cos \phi)+i(\sin n \phi-\sin \phi) \\ & =-2 \sin \frac{n+1}{2} \phi \cdot \sin \frac{n-1}{2} \phi+2 i \sin \frac{n-1}{2} \phi \cdot \cos \frac{n+1}{2} \phi \\ & =-2 \sin \frac{n-1}{2} \phi \cdot\left(\sin \frac{n+1}{2} \phi-i \cos \frac{n+1}{2} \phi\right) \end{aligned}$$

So
$$f(\xi)=\left(\sin \frac{n+1}{2} \phi-i \cos \frac{n+1}{2} \phi\right) \cdot\left(\sin \frac{n+1}{2} \phi-a \sin \frac{n-1}{2} \phi\right)$$

Let $g(\phi)=\sin \frac{n+1}{2} \phi-a \sin \frac{n-1}{2} \phi$. We mainly consider the roots of $g(\phi)=0$ in $[0,2 \pi)$. First, $g(0)=0$, and we discuss the following cases:
1. If $0<a<1$, then $g(\pi)=-\sin \frac{n-1}{2} \pi<0$. Since $g$ is a continuous function, $g$ has at least one root in $\left[\psi_{k}, \psi_{k+1}\right]$. Adding the root $g(0)=0$, $g$ has $n+1$ roots in $[0,2 \pi)$, which means $f$ has $n+1$ roots on the unit circle. However, by the Fundamental Theorem of Algebra, $f$ has only $n+1$ roots in the entire complex plane, so all roots of $f$ are on the unit circle.
2. If $a=1$, then $f(z)=(z-1)\left(z^{n}+1\right)$, and all its roots are on the unit circle.
3. If $1<a$, then $g(\pi)=-a \sin \frac{n-1}{2} \pi<0$. So $g$ has at least one root in $\left[\psi_{k}, \psi_{k+1}\right]$. Adding the root $g(0)=0$, we need to find two more roots.
Since $g(0)=0, g^{\prime}(0)=(n+1)-a(n-1)>0$, but $g\left(\psi_{1}\right)<0$, $g$ has a root in $\left(0, \psi_{1}\right)$. Similarly, $g\left(\psi_{n-1}\right)<0$, but $(-1)^{n-1} g\left(\psi_{n-1}\right)>0$, so $g$ has a root in $\left(\psi_{n-1}, 2 \pi\right)$. Therefore, $g$ has $n+1$ roots in $[0,2 \pi)$, which means $f$ has $n+1$ roots on the unit circle. So, by the same reasoning, all roots of $f$ are on the unit circle.