A-stability of Heun method for ODEs I'm trying to determine the stability region of the Heun method for ODEs by using the equation $y' = ky$, where $k$ is a complex number, based on the method described here.
If the Heun method is: 
$$y_{n+1} = y_n + 0.5\cdot h\bigl(f(t_n, y_n) + f(t_{n+1},y_n + 0.5\cdot h\cdot f(t_n, y_n)\bigr)$$
then when I insert $y' = zy$ for $f(t,y)$, my result simplifies to
$$ y_{n+1} = (0.25\cdot h^2 \cdot z^2 + hz + 1)y_n $$
to judge from the wiki article, the stability region is then the area described by 
$$\\{z \in \mathbb C \mid 0.25h^2z^2 + hz + 1 < 1\\}$$
Am I doing this right? What would such a region look like? Can someone help me get the intuition for this? And then I guess the method is A-stable if that region includes wherever $\Re < 0$?
 A: For the method that you wrote down, you indeed have
$$ y_{n+1} = (0.25\cdot h^2 \cdot z^2 + hz + 1)y_n. $$
The expression between parentheses is a function of $w = hz$, and the stability region consists of the numbers $w$ for which the modulus of the expression between parentheses is at most one:
$$ \text{stability region} = \{ w \in \mathbb{C} : |0.25w^2 + w + 1| < 1 \}. $$
So the stability region does not depend on the step size $h$.
In the particular case you are asking about, you can use that
$$ 0.25w^2 + w + 1 = (0.5w+1)^2. $$
You should be able to find the region from this.
You are right about how to determine whether the method is A-stable. You should find that the method is not A-stable, because explicit methods are never A-stable (see also the Wikipedia page that you link to).
In general, to get a feeling for what the stability region looks like, one may start by restricting to the real axis. If $w$ is real, then $0.25w^2+w+1$ is also real, so the condition $|0.25w^2+w+1| < 1$ simplifies to $-1 < 0.25w^2+w+1 < 1$. But what people often do in practice is to plot the region using the computer.
A final note: Are you sure you copied the method correctly? The method
$$y_{n+1} = y_n + 0.5\cdot h\bigl(f(t_n, y_n) + f(t_{n+1},y_n + h\cdot f(t_n, y_n)\bigr)$$
with no factor 0.5 in front of the last $h$, is more popular.
A: I think you mean to type(?)
$$\{z \in \mathbb C : |0.25h^2z^2 + hz + 1| < 1\}$$
Why to impose this condition? Consider the following at time step $n$:
$$
y_{n} = \Big(\frac{h^2 z^2}{4} + hz + 1\Big)^n \,y_0
$$
therefore having the first stability condition is the sufficient condition for a small disturbance in the initial value wouldn't get magnified marching in the time step: Consider $y_0^{\epsilon} = y_0 + \epsilon$, then
$$
|y^{\epsilon}_{n} - y_n| = \Big\vert\frac{h^2 z^2}{4} + hz + 1\Big\vert^n \,\epsilon < \epsilon
$$
and $\epsilon$ can be happen at any time-step, or even could be the numerical error itself, this is why sometimes for a relatively not-so-small step-size $h$, after a few iterations, the numerical error gets magnified and the numerical solution blows up. 
As for question about A-stableness, based on wiki, I am guessing the answer is Yes.
