Is there an easy way to find the complex roots of this polynomial?

Question

I want to find the complex roots of $1-x-x^2-...-x^k$ for some $k\in\mathbb{N}$. Alternatively, since $(1-x)(1-x-...-x^k) = 1-2x+x^{k+1}$, finding the roots for $1-2x+x^{k+1}$ would also suffice.

Answer 1

It seems likely that we'll need to use numeric methods to approximate the roots, because it isn't obvious that there's a natural factorisation for general values of $k$. However, we can make the following observation:

Let $p(x) = 1 - 2x + x^{k+1} = (1 - x)(1 - x - x^2 - \ldots - x^k)$ (you had a sign error in your question). Then $p'(x) = -2 + (k+1) x^k$. Then if we set $p'(x) = 0$ we get $x^k = \frac{2}{k+1}$. If $k$ is odd then this has a single solution $x = \sqrt[k]{\frac{2}{k+1}}$, while if $k$ is even then there are two solutions $x = \pm \sqrt[k]{\frac{2}{k+1}}$.

From the values of $p(x)$ and $p''(x)$ at these values, we can determine that for $k > 1$ we will always have two roots when $k$ is odd, and three when it's even. But we want to drop the $x = 1$ root, so that means that $1 - x - \ldots - x^k$ will always have 1 root when $k$ is odd, and 2 when $k$ is even. One of those roots will be in the interval $(0, \sqrt[k]{\frac{2}{k+1}})$.

Also, if $k$ is even then the other root can be easily confined to the interval $(-2, -1)$, and you can probably get a tighter bound fairly quickly.