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The Runge-Kutta method has $A=\left(\begin{array}{cc}1/4& 1/4-a\\ 1/4+a& 1/4\end{array}\right)$ and $b=(1/2,1/2{)}^T$. Solving the stage equations for $y^{\prime }=\lambda y$ gives the stability functionFor $a\ne 0$, 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 givesThis proves the A-stability of a symmetric two-stage implicit Runge-Kutta family:

For the test equation $f(y)=\lambda y$, put $z=h\lambda$. The stage equations of this implicit Runge-Kutta method areSubstitution into the update gives the stability functionFor $z=x+iy$, direct expansion givesThis is nonnegative whenever $x\le 0$, and hence $|R(z)|\le 1$ throughout the closed left half-plane, provided the denominator has no zero there.

If $a=0$, the denominator vanishes only at $z=2$. If $a\ne 0$, its zeros solveReal roots are positive, while a complex-conjugate pair has positive real part because the sum of the roots is $1/(2a^2)>0$. Thus there are no poles in the closed left half-plane for any real $a$. Thereforeis the complete set of A-stable parameters, as summarized by the A-stability of a symmetric two-stage implicit Runge-Kutta family.