I was reading a book and found the following theorem without proof.
Theorem:
Let $\phi(z)=1-\phi_1z-...-\phi_p z^p$. If $\phi(z)\neq0$ for $\lvert z\rvert\leq1$ then there exists $\epsilon > 0$ such that $\frac{1}{\phi(z)}$ has a power series expansion: $\frac{1}{\phi(z)}=\sum_{i=0}^\infty \epsilon_jz^j=\epsilon(z)$ for $\lvert z \lvert\leq 1+\epsilon$.
Anyone knows how to prove it, or at least a reference of where can I look for the proof?
Note that any root $a$ of the polynomial $\phi$, (i.e. a complex number $a$ satisfying $\phi (a) = 0$ must have absolute value strictly greater than $1$, or $|a| > 1$. This is because $\phi (z) \not= 0$ for $|z| \le 1$, by hypothesis. Since there can only be finitely many roots of this polynomial, there is a root $a_0$ such that the absolute value $|a|$ is minimal, i.e. $\phi (a_0) = 0$ and $\phi (a) = 0 \Rightarrow |a| \ge |a_0|$. Note that $|a_0| > 1$ by the argument above, so actually you get $\phi (z) \not= 0$ for all $z$ satisfying $|z| < |a_0| = 1 + \epsilon$. Thus $\frac{1}{\phi (z)}$ is a holomorphic function on the disk $|z| < 1 + \epsilon$, and so has a power series expansion $\sum\limits_{j = 0}^\infty \epsilon_j z^j$ for $|z| < 1 + \epsilon$.