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For the test equation $y^{\prime }=\lambda y$, put $z=h\lambda$. The amplification roots satisfyThe quadratic Schur stability criterion shows that both roots lie in the closed unit disk whenTo check these inequalities on the closed left half-plane, write $z=-u+iv$ with $u\ge 0$. The right side of the second inequality isAfter squaring, the difference between its square and $16|5\overline{D}+1+z{|}^2$ is a polynomial in $u$ and $v^2$ with nonnegative coefficients:The first inequality follows from the same displayed positive expression. The unit-circle cases are semisimple, so the entire closed left half-plane is in the linear stability domain. Hence the method is A-stable.

The linear stability domain is $\{z\in \mathbb{C}:|R(z)|\le 1\}$ for the test equation $y^{\prime }=\lambda y$, $z=h\lambda$. A method is A-stable when this domain contains the entire closed left half-plane.