Let $U$ be the exterior of the unit disk in $\mathbb{C}$. That is, $U = \{ z \in \mathbb{C} : |z|>1 \}$. Suppose that the series $f(z)=z+a_0+a_1/z + a_2/z^2 + \cdots$ converges in $U$ where $a_n \in \mathbb{C}$. Are there any sufficient conditions known (in terms of the coefficients $a_n$) for $f$ to be univalent(one-to-one)?
It will be particularly nice if the condition(s) are given in terms of the decay rate of $a_n$.
Thanks in advance.
Just to simplify the calculus let $$g(z)=f(1/z)=\frac1z+\sum_{n=0}^\infty a_nz^n$$and note that $f$ is injective in $\{|z|>1\}$ if and only if $g$ is injective in the punctured unit disc $D=\{0<|z|<1\}$.
First note that $$\left|\frac1z-\frac1w\right|=\frac{|z-w|}{|zw|}\ge|z-w|\quad(z,w\in D).$$
Since $$\left|\frac d{dz}z^n\right|\le n\quad(n\in\Bbb N,|z|<1)$$it follows that $$|z^n-w^n|\le n|z-w|\quad(n\in\Bbb N,z,w\in D).$$Hence $$|g(z)-g(w)|\ge\left( 1-\sum_{n=1}^\infty n|a_n|\right)|z-w|\quad(z,w\in D),$$so if $$\sum_1^\infty n|a_n|<1$$ then $g$ is injective in $D$.