Polynomial in $z$ not bounded on $\mathbb{C}^-$

Question

This is part of a proof that shows that explicit Runge-Kutta-methods are not A-stable. It states the following: Given we have a polynomial $R(z)\in P_s$ in $z$ of degree at most $s\geq p>0$, then $R(z)$ is not bound on the set $$ \mathbb{C}^-:= \{z\in \mathbb{C}: Re(z)\leq0 \} $$ This was stated without a proof or explanation and is the only part in the proof I am struggeling with, so I would appreciate your help.

Answer 1

I'm assuming $s>0$, as $0$ order methods don't make sense as explained to me in the comments. Polynomials are entire (holomorphic on the complex plane). So by Liouville's theorem, which states that a bounded entire complex function must be constant, $R(z)$ your polynomial must be unbounded on $\mathbb{C}$. To show it is unbounded on $\mathbb{C}^-$ you can first work with a monic polynomial and note it is unbounded on $\mathbb{C}^-$ by symmetry. To see this, suppose $z^m$ it is unbounded on just the right half plane. Then $(re^{i\theta})^m = r^m (\cos(m\theta)+i\sin(m\theta))$ is unbounded on the right half plane. Then reflecting across the imaginary axis yields $r^m(\cos(m\pi-m\theta)+i\sin(m\pi-m\theta))$ is also unbounded.

Since a polynomial $R$ of degree $s$ is such that $\lim_{|z|\to\infty} \frac{z^{s-1}}{R(z)}=0$ then $R(z)$ is unbounded, since $z^{s-1}$ is unbounded by the previous paragraph.

Answer 2

But... $x^s$ is unbounded, isn't it ?