 Solution

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: