Can someone give analytical expressions for the complex roots of the polynomial of the form:
$$(1+x)^n + (1+x)^m - 1,$$
where $n$ and $m$ are arbitrary integers.
Basically, I want an analytical explanation of the following region (which are the roots achieved from mathematica):
I doubt that in general there is a closed-form for the roots in terms of $n$ and $m$, except for the easy cases where $n=m$, or one of them is zero. For an "analytical explanation" of your plot, suppose without loss of generality that $n>m\geq 1$, and make the natural substitution $z=1+x$, so the polynomial is $z^n+z^m-1$. Then for all $\zeta$ such that $\vert \zeta\vert=1.62$, you can check that $\vert \zeta^n\vert>\vert\zeta\vert^m+1\geq \vert\zeta^m-1\vert$. Rouche's Theorem then implies all $n$ roots lie in the disc of radius $1.62$ centered at $0$. Translating back, all the roots of your polynomial are in the disc of radius $1.62$ centered at $x=-1$ in the complex plane. For specific choices of $n,m$ with $n>m$, you can make this tighter.
Also, there's always a real root between $-1$ and $0$ by the intermediate value theorem. There are no positive roots.
Edit: If you want an even better bound on the roots, going back to $z^n+z^m-1$, it follows from Landau's inequality that the product of the magnitudes of the roots with magnitude at least $1$ here is bounded by $\sqrt{3}$. Any nonreal root will come with a conjugate pair with the same norm, so each must have magnitude at most $3^{1/4}\approx 1.32$. You can use Descartes' Law of Signs and probably the intermediate value theorem to show there are only a couple of real roots, and they are pretty close to $0$. So Landau's inequality gives a pretty good radius on almost all of the roots.