Suppose that \(f\) is holomorphic in an open set containing the closed unit disc, except for a pole at \(z_0\) on the unit circle. Let \(\displaystyle %f(z) = \sum_{n = 1}^\infty a_n z^n f(z) = \sum_{n = 1}^\infty c_n z^n\) denote the the power series in the open disc. Show that (1) \(c_n \neq 0\) for all large enough \(n\)’s, and (2) \(\displaystyle \lim_{n \rightarrow \infty} \frac{c_n}{c_{n+1}}= z_0\).
Solution
Let the pole of \(f\) at \(z_0\) have order \(m\), with principal part \[ \sum_{j=1}^m\frac{A_j}{(z-z_0)^j}, \qquad A_m\ne0. \] Because \(f\) is holomorphic on an open neighborhood of the closed unit disk except at \(z_0\), there is some \(R>1\) such that after subtracting this principal part the remainder \[ h(z)=f(z)-\sum_{j=1}^m\frac{A_j}{(z-z_0)^j} \] is holomorphic on \(|z|<R\).
For \(|z|<1=|z_0|\), \[ \frac1{(z-z_0)^j} =(-1)^j z_0^{-j} \sum_{n=0}^\infty \binom{n+j-1}{j-1}\frac{z^n}{z_0^n}. \] Hence the coefficient \(c_n\) of \(z^n\) has the form \[ c_n =z_0^{-n}\sum_{j=1}^m (-1)^jA_jz_0^{-j}\binom{n+j-1}{j-1} +d_n, \] where the Taylor coefficients of \(h\) satisfy \(d_n=O(R^{-n})\) by Cauchy’s estimates.
The \(j=m\) term has degree \(m-1\) in \(n\) and nonzero leading coefficient, whereas all terms with \(j<m\) have smaller polynomial degree. Therefore \[ c_n=Cz_0^{-n}n^{m-1}\left(1+O\left(\frac1n\right)\right) \] for some constant \(C\ne0\); the exponentially small term \(d_n\) does not affect this asymptotic. Consequently \(c_n\ne0\) for all sufficiently large \(n\), and \[ \frac{c_n}{c_{n+1}} =z_0\left(\frac{n}{n+1}\right)^{m-1} \frac{1+O(1/n)}{1+O(1/n)} \longrightarrow z_0. \] This proves both assertions.