I'm working on the following problem concerning Rouché's theorem:
Let $n\geq2.$ Show that the polynomial $f(z)=z^n+z+1=0$ has $n$ distinct complex roots in $D_2=\{z\in\mathbb{C}\hspace{0.1cm}|\hspace{0.1cm}|z|\leq2\}.$
I am comfortable showing the "$n$ roots" part using Rouché's theorem, picking $g(z)=z^n$ and $h(z)=z+1.$ However, I have trouble with how to approach showing the distinctness of these $n$ roots.
Is it recommended to show that ${\text{gcd}}(f,f')=c,$ a constant, implying that $f$ has no multiple root? (If $f$ had a multiple root at $z=a,$ then writing $f(z)=(z-a)^2\cdot k(z)$ and taking the derivative of both sides shows that $z=a$ is a root of $f'$ as well so the gcd couldn't be degree $0.$)
However, it's not immediately obvious to me that for $n\geq2$ we have ${\text{gcd}}(z^n+z+1,nz^{n-1}+1)=c.$
Is there another way I should be approaching this? Any guidance is appreciated.
Thank you.
Suppose that $z^n+z+1$ and $nz^{n-1}+1$ have a common root $w$. Then $w^{n-1}=-\frac1n$ and therefore $w^n=-\frac wn$. But $w^n=-w-1$, and so $-\frac wn=-w-1$. Therefore , $w=\frac n{1-n}$. But, clearly, $\frac n{1-n}$ is not a root of $nz^{n-1}+1$.