Runge Kutta stability region for forward Euler and explicit midpoint The interval of absolute stability is the intersection of the region of absolute stability in the complex plane with the real axis.Show that Runge Kutta forward Euler and RK explicit midpoint have the same interval of absolute stability :
I know that the stability function :
$R(z) = 1 +h\lambda$ for RK forward Euler
and
$R(z) = 1+h\lambda+(\lambda*h)^2/2$ for RK explicit midpoint .
also some text show that $R(z)=\exp(i\theta)$ for $\theta$ in $[0,2\pi]$
I don't know much about complex analysis so please help.
 A: The criterion for absolute stability is that
\begin{align}
1 & \overset{!}{\geq} \vert R(z) \vert  \\
\Leftrightarrow 1 & \overset{!}{\geq} \vert R(z) \vert^2  \\
\Leftrightarrow 1 & \overset{!}{\geq} \text{Re}^2\big(R(z)\big)+ \text{Im}^2\big(R(z)\big) 
\end{align}
Since we are just interested in the maximum interval of stability, we take $$z := h \lambda \in \mathbb R$$
and thus $$ \text{Im}\big(R(z)\big)  \equiv 0 $$
and $$\text{Re}\big(R(z)\big) \equiv R(z) $$
since $R(z)$ has real coefficients.
For Forward Euler, the criterion becomes
$$ \vert 1 + z \vert \overset{!}{\leq} 1 $$
where the minimum value for $z \in \mathbb R^-$ is obviously $z_\text{min} = -2$.
Now, for explicit midpoint we have to do some more algebra.
\begin{align}
1 & \overset{!}{\geq} \vert 1 + z + 0.5 z^2 \vert  \\ 
\end{align}
You can now plot $ 1 + z + 0.5 z^2$ and observe that it is for real $z$ always positive.
As a consequence, we can drop the absolute value and solve at the stability boundary
\begin{align}
1 &= 0.5z^2 + z + 1 \\
\Leftrightarrow 0 &=  z(0.5z + 1) 
\end{align}
which has the solutions $z=0$ and $z=-2$.
