Coefficients of this Taylor series are bounded

Question

Let $f(z)$ be a function that's analytic in the open unit disk, and also in a region containing the closed unit disk with the exception of a few simple poles (i.e. the poles have degree 1) that lie outside the open unit disk. Show that the coefficients of the Taylor expansion of f(z) at 0 (we know its radius is at least the unit disk) are bounded.

I could use help with this. What I've tried: we can multiply $f(z)$ by $(z-c_1)(z-c_2)...$ where $c_1,...$ are the singularities to get an analytic function, and then express the series of $(z-c_1)(z-c_2)...f(z)$ in terms of $f(z)$'s series to get a constraint on the coefficients (since the analytic function's coefficients will go to 0). But this is tedious and the constraint doesn't seem good enough to show boundedness.

Update: Sorry, it looks like I confused a few people (as seen by the hints below). To clarify, $f$ is analytic in the open unit disk. We're not given analyticity in the closed unit disk.

I should also emphasize that the poles are of degree 1 at the most.

This isn't a homework question (just an exercise from a textbook), so feel free to post either hints or whole answers.

Answer 1

The function $f$ can be written as

$$ f(z)=g(z)+\sum_{k=1}^N\frac{a_k}{z-z_k} $$

for some $r>1$ and $|z|<r$. Here $g$ is holomorphic and $|z_k|\geq 1$ for $k\in\{1, \dotsc, N\}$. The boundedness of the coefficients follows directly from this decomposition (note that the radius of convergence of $g$ is at least $r>1$).

Answer 2

The function $\displaystyle f(z)=\frac1{(1-z)^2}$ has Taylor series $\displaystyle \sum_{n=0}^\infty (n+1)z^{n}$, a double pole at $z=1$, and unbounded coefficients.

You mentioned that the poles of $f$ are simple and lie outside the open unit disc. The example above shows that if we omit the requirement that the poles are simple, then we actually need the poles not to be in the boundary. But then, for such an $f$, by an easy compactness argument, $f$ is actually analytic in a slightly larger open disc: Cover the boundary with small balls where $f$ is analytic. These exist since the poles lie outside the closed disc. Use compactness to reduce to a finite covering. This gives us an open set containing the closed disc where $f$ is analytic, by the uniqueness theorem. Any such set contains an open disc centered at the origin, and of radius larger than $1$.

Say $\displaystyle f(z)=\sum_{n=0}^\infty a_n z^n$. For $z$ of size $1$, the above implies that series converges, so $$|a_n|=|a_nz^n|\to 0.$$ This implies boundedness.

(Of course, this does not address the case we are interested in, where the poles are simple and may lie in the boundary.)

To address this case: By compactness, for some $\epsilon>0$, there are finitely many poles of $f$ in the disc centered at $0$ or radius $1+\epsilon$ (and they are all simple). The point is that we can pick $\epsilon$ small enough, so $f$ is meromorphic in the circle of radius $1+2\epsilon$, say. If there are infinitely many poles in the disc of radius $1+\epsilon$, they have an accumulation point, and at that point $f$ is not analytic and cannot have a pole.

The problem is easy now: If $a$ is one of these poles, then $\displaystyle f(z)=g(z)+\frac{c}{z-a}$ where $g$ is analytic where $f$ is, and also at $a$. Applying this finitely many times, we conclude that $$ f(z)=h(z)+\sum_{k\le n}\frac{c_k}{z-a_k} $$ where $h$ is analytic in $|z|\le 1$, the $a_k$ are the simple poles of $f$ in the disc of radius $1+\epsilon$, and the $c_k$ are constant. Combining this with the argument in the third paragraph proves the result, as $(z-a_k)^{-1}$ can be expanded directly as a geometric series and, since the $a_k$ have modulus at least $1$, the coefficients of these geometric series are bounded.