Suppose $f : \mathbb{C} \to \mathbb{C}$ is a polynomial such that $\forall z \in \mathbb{C}: |zf(z)| \leq |z|^k$. Does it follow that $ \forall z \in \mathbb{C}: |f(z)| \leq |z|^{k-1}$ ?
My thoughts: Certainly this holds $\forall z \neq 0$. For $z = 0 $, I believe this should also be true by continuity of $f$. So I would say the assertion is valid.
However I am unsure. I would be grateful if someone could either verify this for me or point out where I am mistaken!
To get the question out of the unanswered queue, note that: $$\forall n \in \mathbb Z_{>0}, f\left(\frac{1}{n}\right) < \frac{1}{n^{k-1}}. $$
It follows that $$ \lim_{n \to \infty} f\left(\frac{1}{n}\right) = 0. $$
By continuity of $f$, we conclude that $f(0) = 0$.