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Consistency was established in part (a). By the Dahlquist equivalence theorem, convergence additionally requires the root condition for a multistep method. Since $\rho (w)=(w-1)(w-\alpha )$, its roots lie in the closed unit disk with each unit-modulus root simple exactly forThe endpoint $\alpha =-1$ is zero-stability although it has a simple parasitic root at minus one; $\alpha =1$ has a repeated root at one and is excluded.

For every parameter in this convergent range, the leading coefficient of $\sigma$ is $(5+\alpha )/12>0$, whileSince $\sigma (w)\to +\infty$ as $w\to -\infty$, it has a real root $w_{\ast }<-1$. Its other real root lies between minus one and one because $\sigma (1)=1-\alpha >0$, so the exterior root is simple.

On the Dahlquist test equation, the amplification polynomial of a multistep method is $P_z(w)=\rho (w)-z\sigma (w)$, with $z=h\lambda$. For $|z|\to \infty$, the equivalent polynomial $-P_z/z=\sigma -\rho /z$ tends coefficientwise to $\sigma$, and its leading coefficient stays nonzero. Choose a small closed disk about $w_{\ast }$ lying wholly outside the unit disk and containing no other root of $\sigma$. On its boundary $|\sigma |$ has a positive minimum, while $|\rho /z|$ is smaller for every sufficiently large $|z|$. Rouché's theorem then puts one root of $P_z$ inside this exterior disk for all such complex $z$.

That root violates the stability root condition for a multistep method. Hence there is a finite radius $R_{\alpha }$ beyond which no point belongs to the absolute stability domain:This proves boundedness in every complex direction, rather than only excluding A-stability. It is an instance of exterior roots of the derivative polynomial bound multistep stability.