= Solution
{c}
As appropriate for a genuine $k$-step <linear multistep method>, take $\alpha_k\ne0$. Suppose the <characteristic polynomial> has a <polynomial root> $\xi$ with <modulus> $|\xi|>1$. From $\rho(\xi)=0$ and the <triangle inequality>,
$$
\begin{aligned}
|\alpha_k||\xi|^k
&=\left|\sum_{j=0}^{k-1}\alpha_j\xi^j\right|\\
&\leq\sum_{j=0}^{k-1}|\alpha_j||\xi|^j\\
&\leq|\xi|^{k-1}\sum_{j=0}^{k-1}|\alpha_j|\\
&\leq|\alpha_k||\xi|^{k-1}.
\end{aligned}
$$
Dividing by the positive number $|\alpha_k||\xi|^{k-1}$ gives $|\xi|\leq1$, a contradiction. Hence
$$
\boxed{|\xi_i|\leq1\qquad(i=1,\ldots,k).}
$$
The nonzero leading <coefficient> matters: if $\alpha_k=0$, the assumed inequality forces all the $\alpha_j$ to vanish, so $\rho$ is the zero <polynomial> rather than a <characteristic polynomial> with $k$ roots. This degenerate case does not define the method in the question.
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