Laurent Series with negative terms

Question

Let $f(z)$ be a complex polynomial sucha that $f(z)\neq 0$ for all $z$ such that $|z| \ge 1$. This implies that there exists $r>0$ such that $1 \over f(z)$ is holomorphic (complex differentiable) in the domain: $ r <|z| < \infty$ and $ r < 1$.

My question: Can we guarantee the exitence of a Laurent series centered at $0$ such that the positive terms vanish? That is, can we guarantee a Laurent Series of the form :

$${1\over f(z)} = \sum_{i = 1}^{\infty}b_{i}z^{-i}$$ such that the series is absolutely convergent for all $z$ in the domain $ r <|z| < \infty$

Any comments or hints would be highly appreciated

Answer 1

You are asking if $\frac 1 {f(\frac 1 z)}$ has a power series expansion in $|z| <\frac 1 r$. This is true because any non-constant polynomial has the property $|f(z) | \to \infty$ as $|z| \to \infty$ so $\frac 1 {f(\frac 1 z)}$ has a removable singularity at $0$ which imples that it has an expansion of the form $ \sum\limits_{i=0}^{\infty} b_iz^{i}$ for $|z| <\frac 1 r$. Note that $b_0=0$.

Answer 2

I believe the answer is yes. By a partial fractions decomposition, $1/f(z)$ can be written as a linear combination of terms of the form $1/(z-z_0)^k$ with $|z_0|<1$, and so it suffices to show that any such function has such a Laurent series; but that follows just from the geometric series and its derivatives.