This is my first question in mathSE, hope that it is suitable here!
I'm currently self-studying complex analysis using the book by Stein & Shakarchi, and this is one of the exercises (p.67, Q14) that I have no idea where to start.
Suppose $f$ is holomorphic in an open set $\Omega$ that contains the closed unit disc, except for a pole at $z_0$ on the unit circle. Show that if $f$ has the power series expansion $\sum_{n=0}^\infty a_n z^n$ in the open unit disc, then
$\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0$.
If the limit is taking on $|\frac{a_n}{a_{n+1}}|$ and assume the limit exists, by the radius of convergence we know that the answer is $1$. But what can we say about the limit of the coefficient ratio, which is a pure complex number? I've tried to expand the limit directly by definition, with no luck. And I couldn't see how we can apply any of the standard theorems in complex analysis.
I hope to get some initial directions about how we can start thinking on the problem, rather than a full answer. Thank you for the help!
Best Answer
Hint: Assume that you have a simple pole at $z = z_0$, where $|z_0| = 1$ and try to prove it. In particular, take $f(z) = \frac{g(z)}{z-z_0}$ where $g(z)$ is holomorphic on $\Omega$. Prove the result for this case. (Expand $\frac{1}{z-z_0}$ about $z=0$ and do some manipulations). Now the same idea can be extended for higher order poles.
EDIT: For a simple pole, $f(z) = \frac{g(z)}{z-z_0} = \displaystyle \sum_{n=0}^{\infty} a_n z^n$. Since $g(z)$ is holomorphic, $g(z) = \displaystyle \sum_{n=0}^{\infty} b_n z^n$. So $\displaystyle \sum_{n=0}^{\infty} b_n z^n = (z-z_0) \displaystyle \sum_{n=0}^{\infty} a_n z^n \Rightarrow b_{n+1} = a_n - z_0 a_{n+1}$.
Now what can we say about $\displaystyle \lim_{n \rightarrow \infty} b_n$ and $\displaystyle \lim_{n \rightarrow \infty} a_n$?
(Note: $g(z)$ holomorphic on $\Omega$ whereas $f(z)$ is holomorphic except at $z_0$, a point on the unit disc).
This same idea will work for higher order poles as well.