Let's consider such polynomial: $$ W_n(z)=\sum_{k=1}^n kz^{k-1} $$ Tell whether this statement is true for any $n \in \mathbb{N}$: $$W_n(z)=0 \Rightarrow |z|<1$$
Here is what I have evaluated:
It is obvious that $z$ can't be a real number bigger or equal to $0$. Also, if $n$ is even, then $z$ can't be a real number smaller than $-1$, because if it was, we would have: $$W_n(z)=\sum_{k=1}^n kz^{k-1}=\sum_{k=1}^{\frac{n}{2}}\Big((2k+1)z^{2k}+2kz^{2k-1}\Big)=\sum_{k=1}^{\frac{n}{2}}z^{2k-1}\Big(2kz+2k+z\Big)$$ Both $z^{2k-1}$ and $2kz+2k+z$ are smaller than $-1$ so each term of the last sum is bigger than $1$ so the final sum is bigger than $0$.
$W_n$ is the derivative of $P_n(z):= \sum_{k=0}^nz^k$, the roots of which are the $n+1$-th roots of unity (with the exception of $1$). By Gauss-Lucas theorem, the roots of $W_n$ lie in the convex hull of the roots of $P_n$. Besides, the roots of $P_n$ are all distinct, thus have multiplicity $1$ (so they are not roots of $W_n$). Therefore, the roots of $W_n$ lie stricly inside the unit disk, i.e. they have modules $<1$.