Are all non-real complex roots of $f(x)=x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k$ have magnitudes less than 1?

Question

We know the following fact about the roots of the polynomial $f(x)=x^n-kx^{n-1}-kx^{n-2}-\cdots-kx-k$, where $n,k$ are integers and $n,k\geq 2$: if $n$ is odd, then $f$ has a positive root in the open interval (k,k+1) and $n-1$ non-real complex roots; on the other hand, if $n$ is even, then $f$ has a positive root in the open interval (k,k+1), a negative root in the open interval (-1,0), and $n-2$ non-real complex roots.

Numerical results suggest the conjecture that all non-real complex roots of $f$ have magnitudes less than 1, and the problem is how to prove/disprove this conjecture?

Answer 1

By using the geometric series, we can simplify the equation as $$ x + \frac{k}{x^{n}}= k+1 $$ which includes one more zero, $x = 1$. Then we have $$ k+1 \leq |x| + \frac{k}{|x|^{n}} = r + \frac{k}{r^{n}} $$ where $r = |x|$. It is easy to check that the function $f(t) = t + \frac{k}{t^{n}}$ is convex since $$ f'(t) = 1- \frac{kn}{t^{n+1}}, \quad f''(t) = \frac{kn(n+1)}{t^{n+2}} > 0. $$ So $f(t) = k+1$ has two positive zeros, 1 and $\alpha \in (k, k+1)$. Thus for $t>0$, $f(t) \geq k+1$ implies $0<t<1$ or $t>\alpha > k$ (which corresponds to the real root).

Answer 2

Apply the Schur-Cohn test. Let $S_n = \sum_{i = 1}^n x^i$. The first step gives $$T[f] = k (k + 1) \left( S_{n - 1} + 1 -\frac 1 k \right).$$ All the subsequent steps give $$T[C (S_i + a)] = C^2 (a - 1) (S_{i - 1} + a + 1).$$ Therefore $T^2[f](0)$ is negative and all the other $T^i[f](0)$ are positive. The number of roots inside the unit disk is $\sum_{i = 1}^1 (-1)^{i - 1} (n + 1 - \kappa_i) = n - 1$.