Boundedness of Stability Region of Explicit LMMs I see that for an explicit method such as AB-n or Forward Euler etc. that the region of absolute stability is always within some contour. Is there a reason why this is? I have tried to deduce a conclusion from the stability polynomial but have come up short...I know that, $$\pi_k(z)=\rho(z)-k\lambda\sigma(z)$$ and considering the roots here, then $$k\lambda= \frac{\rho(z)}{\sigma(z)}$$
The method is convergent if $\rho(1)=0$, $\rho^{'}(1)=\sigma(1) \neq 0 $, and if the roots of $\rho$ are less than or equal to 1, with $|z_i|=1$ a simple root. Now if that is true we know $\rho$ has a root of 1. So, $$k\lambda=(z-1)f(z)$$ where $f(z)$ is a rational function with degree of numerator = degree of denominator since the method is explicit. Now, $p(z)$ has lead coefficient 1, so the numerator of $f$ does also, but $\sigma$ may or may not so the denominator of $f$ may or may not.
We also know that the value of $z$ is taken such that $|z|=1$ (ie a number from the unit circle). So all factors of the numerator are less than 1 for all $z$. Thus the numerator is less than 1.
So it is known that $|z-1| \leq1$ but is it always true that $f$ is? If so we are golden...
 A: Let $$\rho(S)=\sum_{j=0}^r a_jS^j$$ and let $$\sigma(S)=\sum_{j=0}^r B_jS^j$$ Then we have $$\pi(S;z)=\rho(S)-z\sigma(S)$$ 
If we have $\tilde{S}$ as a root of $\pi(S;z)=\rho(S)-z\sigma(S)$ then it is also a root of 
$$\hat{\pi}(S,z)=\frac{\rho(S)}{z}-\sigma(S)$$
Next, as $z\rightarrow \infty$ we have $$\hat{\pi}(S;z)\approx -\sigma(S)$$
Since the roots of a polynomial vary continuously with the coefficients, we know that $r-1$ out of the r roots of $\hat{\pi}$ will converge to the $r-1$ roots of $-\sigma(S)$ as $z\rightarrow \infty$
Note that $\pi(S;z)$ and $\hat{\pi}(S;z)$ each have $r$ roots. The rth root of $\pi(S;z)$ that hasn't been accounted for is the one we must consider. This root will either converge to a finite value or the root will tend to infinity with $z$. 
If it converges, it must converge to a finite value equal to one of the existing roots of $-\sigma(S)$. But, if that were the case, $-\sigma(S)$ would have r roots which contradicts the idea that it has only $r-1$ roots since it is only degree $r-1$. 
Therefore, we conclude that the root must tend to infinity, and, as a direct consequence, $\pi(S;z)$ will not satisfy the root conditions for $z\rightarrow \infty$. Hence, the set of values of z that satisfy the root conditions must be bounded which gives that the RAS is bounded.
