Properties of function analytic in $\mathbb C$ except for pole at $z=1$, with pole at $\infty$ and no zeros I read a question in some book that if $f$ is analytic in $\mathbb C$ except for a pole at $z=1$, $f$ has a pole at $\infty$, and $f$ has no zeros, what can you say about $f$? 
All I know is that $1/f$ is well defined and $|1/f(z)| \to 0$ as $z \to\infty$ and $z \to 1$. What is special about $f$ in that case?
 A: Since $f$ has a pole at $\infty$, we know that $f$ is a rational function, i.e., $f(z) = \frac {P(z)}{Q(z)}$, where $P$ and $Q$ are polynomials. Moreover, the pole at infinity tells us that  $P$ has a higher degree than $Q$. But since $f$ has no zeros, this means that $P$ has no zeros, and so $P$ has degree 0. But on the other hand, since $Q$ must have a zero at $z=1$, we know that the degree of $Q$ is at least 1. This is a contradiction (in order to have a pole at infinity, we need deg $P$ > deg $Q$), and tells  us that $f$ cannot exist!
A: As you say, $g(z) = 1/f(z)$ is well defined on $\Bbb C - \{1\}$ since $f$ has no zeros. Also, $\left|g(z)\right| \to 0$ as $z \to 1$, so $g$ is bounded in a deleted neighborhood of $1$. It follows that $1$ is a removable singularity of $g$.
We have $\left|g(z)\right| \to 0$ as $|z| \to \infty$. From this we can find $M$ so that $\left|g(z)\right| < 1$ when $|z| > M$. $\left|g(z)\right|$ is bounded on $|z| < M$ because it's entire. It follows that $g$ is entire and bounded, hence constant by Liouville's theorem.
Thus, $f$ is also constant. This contradicts with the assumptions. We conclude that no such $f$ exists.
