= Solution
The <Runge-Kutta method> has $A=\begin{pmatrix}1/4&1/4-a\\1/4+a&1/4\end{pmatrix}$ and $b=(1/2,1/2)^T$. Solving the stage equations for $y'=\lambda y$ gives the <stability function>
$$
R(z)=1+z\,b^T(I-zA)^{-1}\mathbf1
=\frac{1+z/2+a^2z^2}{1-z/2+a^2z^2}.
$$
For $a\ne0$, the denominator roots have positive real parts: real roots are positive from their positive sum and product, while a complex conjugate pair has real part $1/(4a^2)>0$. For $a=0$ its sole root is $2$. Thus there are no poles or singular stage systems in the left half-plane.
Writing $z=x+iy$, direct expansion gives
$$
|1-z/2+a^2z^2|^2-|1+z/2+a^2z^2|^2=-2x(1+a^2|z|^2)\ge0\quad(x\le0).
$$
This proves the <A-stability of a symmetric two-stage implicit Runge-Kutta family>:
$$
\boxed{\text{The method is A-stable for every }a\in\mathbb R.}
$$
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