= Solution
For $s=2$, part (a) gives $\rho(w)=\tfrac32w^2-2w+\tfrac12$ and the right-side <polynomial> is $w^2$. On the scalar test equation $y'=\lambda y$, put $z=h\lambda$. Its characteristic equation is
$$
(\tfrac32-z)w^2-2w+\tfrac12=0,
\qquad w_\pm=\frac{2\pm\sqrt{1+2z}}{3-2z}.
$$
The <linear stability domain> contains every $z$ for which both roots have modulus strictly less than one. For real $-1/2\le z<0$, the roots are positive; their larger numerator is less than three while their denominator exceeds three, so both are less than one. At $z=-1/2$ the repeated root is $1/2$, also strictly inside the <unit circle>. For $z<-1/2$, the roots are <complex conjugates> and their product is
$$
|w_\pm|^2=\frac1{3-2z}<1.
$$
Thus in every case both roots lie strictly inside the <unit disk>. Repeated roots there are allowed, since <polynomial> factors times their powers still tend to zero. Therefore
$$
\boxed{(-\infty,0)\text{ lies in the linear stability domain of the two-step BDF method}.}
$$
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