Complex Analysis Question from Stein

The question is #$14$ from Chapter $2$ in Stein and Shakarchi’s text Complex Analysis:

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. Show that if $$\sum_{n=0}^\infty a_nz^n$$ denotes the power series expansion $f$ in the open unit disc, then $$\lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0.$$

I’ve shown that we can take $z_0=1$ without a loss of generality, but I’m having trouble showing the proof otherwise. One of the problems I’m having is because we aren’t told the definition of a pole except that it is a place where the function isn’t holomorphic. Disregarding this fact, the other problem I’m running into is that I don’t know the order of the pole.

Making some additional assumptions, including that the pole is simple so we can write $F(z)=(z-1)f(z)$ as a holomorphic function, we see that $$F(z)=-a_0+z(a_0-a_1)+z^2(a_1-a_2)+\cdots$$ This almost gets me to the end with these added assumptions, but I don’t think it’s quite enough (why do we know some of the $a_i$’s aren’t $0$, for example).

On another note, if we know $\lim_{n\to\infty}\frac{a_n}{a_{n+1}}$ exists, then it is easy to see that $\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}=1=|z_0|$; I, however, do not see why the limit must exist.

Are there any hints that someone can provide? Even a solution would be nice, especially if one can avoid making any assumptions about what a pole is or is not.

EDIT: So there isn’t any confusion, I know the definition of a pole and I’m inclined to believe that the problem, as stated, necessarily has a pole at $z_0$. The problem is that the exercise is in Chapter $2$, and poles are introduced in Chapter $3$.

Solutions Collecting From Web of "Complex Analysis Question from Stein"

I will assume we can write

$$f(z) = \frac{c}{z_0-z} + \sum_{n=0}^{\infty} b_n z^n$$

for some value of $c \ne 0$, and $\lim_{n \to \infty} b_n = 0$. Then

$$f(z) = \sum_{n=0}^{\infty} a_n z^n$$


$$a_n = b_n + \frac{c}{z_0^{n+1}}$$


$$\begin{align}\lim_{n \to \infty} \frac{a_n}{a_{n+1}} &= \lim_{n \to \infty} \frac{\displaystyle b_n + \frac{c}{z_0^{n+1}}}{\displaystyle b_{n+1}+ \frac{c}{z_0^{n+2}}}\\ &= \lim_{n \to \infty} \frac{\displaystyle \frac{c}{z_0^{n+1}}}{\displaystyle \frac{c}{z_0^{n+2}}}\\ &= z_0\end{align}$$

as was to be shown. Note that the second step above is valid because $z_0$ is on the unit circle.

For a nonsimple pole, we may write

$$f(z) = \frac{c}{(z_0-z)^m} + \sum_{n=0}^{\infty} b_n z^n$$

for $m \in \mathbb{N}$. It might be known that

$$(1-w)^{-m} = \sum_{n=0}^{\infty} \binom{m-1+n}{m-1} w^n$$


$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{n+1}{n+m} z_0$$


@TCL observed that we can simply require that $b_n z_0^n$ goes to zero as $n \to \infty$. Then for a simple pole

$$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim_{n \to \infty} \frac{\displaystyle b_n z_0^n + \frac{c}{z_0}}{\displaystyle b_{n+1} z_0^n + \frac{c}{z_0^2}}$$

which you can see goes to $z_0$.

This is just an attempt to give a rigorous proof of Ron Gordon’s answer.

Suppose the order of the pole at $z_0$ is $m$. Then for some constants $c_0,\ldots,c_m, c_m\neq 0,$
$$g(z)=f(z)-\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$
is analytic at $z_0$. Since $f(z)$ is analytic on an open set containing the unit disk except at $z_0$, we see that $g(z)$ is analytic on an (possibly different) open set containing the unit disk. So $$g(z)=\sum_{n=0}^\infty b_nz^n$$ has radius of convergence greater than 1, in particular the series converges at $z_0$ and $\lim_n b_nz_0^n=0$. So for $|z|<1$,
$$f(z)=\sum_{n=0}^\infty a_nz^n=\sum_{n=0}^\infty b_nz^n+\left(\frac{c_m}{(z_0-z)^m}+\cdots +\frac{c_1}{(z_0-z)}\right)$$
For $1\le k\le m$, we have $$\frac{c_k}{(z_0-z)^k}=\sum_{n=0}^\infty \frac{(k)_nc_k}{n!z_0^{n+k}} z^n$$
where $(k)_n=k(k+1)\cdots(k+n-1)$ and $(k)_0=1$. It follows that $$a_n=b_n+\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{n+k}}$$ and
\begin{align*}\frac{a_n}{a_{n+1}}&=\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{1+k}}+b_{n+1}z_0^n}\\
&= z_0 \frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}}
Now divide the numerator and denominator of the last expression by $$\frac{(m)_{n+1}}{(n+1)!}$$, which is equal to 1 when $m=1$ and approaches infinity as $n\to\infty$ if $m>1$. Furthermore, for $1\le k\le m$,
$$\frac{(k)_n}{n!}\cdot \frac{(n+1)!}{(m)_{n+1}}=\frac{(k)_n (n+1)}{(m)_n (n+m)}$$
approaches 0 as $n\to\infty$ if $k<m$ and approaches 1 if $k=m$. It follows that the fraction
$$\frac{\sum_{k=1}^m \frac{(k)_nc_k}{n!z_0^{k}} +b_nz_0^n}{\sum_{k=1}^m \frac{(k)_{n+1}c_k}{(n+1)!z_0^{k}}+b_{n+1}z_0^{n+1}}$$
approaches 1 as $n\to\infty$, and the proof is complete.

EDIT. If $k<m$, $$\frac{(k)_n }{(m)_n }=\prod_{i=0}^{n-1} \frac{k+i}{m+i}=\prod_{i=0}^{n-1} \left(1-\frac{m-k}{m+i}\right)\to 0$$ because $\sum_{i=0}^\infty \frac{m-k}{m+i}=\infty$.

Suppose that
&=(z-z_0)^n\sum_{k=0}^\infty a_kz^k\\
&=\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}z^jz_0^{n-j}\sum_{k=0}^\infty a_kz^k\\
converges in a neighborhood of $z_0$. Thus, for some $r\lt1$, we have
\left|\sum_{j=0}^n(-1)^{n-j}\binom{n}{j}a_{k-j}z_0^{n-j}\right|\le c\,r^k
That is, the the $n^{\text{th}}$ finite difference of the sequence $a_kz_0^k$ satisfies
Inverting the finite difference operator yields that there is an $n-1$ degree polynomial $P$ so that
Taking the limit of the ratio of terms yields
That is,

Inverting Finite Difference Operators

If we define the finite difference operator as
\Delta a_k=a_k-a_{k-1}
then, as with indefinite integrals, when inverting $\Delta$, we need to include a constant:
\Delta^{-1}a_k=c+\sum_{j=1}^k a_j
Suppose that $\Delta a_k=P_n(k)+O(r^k)$, where $P_n$ is a degree $n$ polynomial and $0\le r\lt1$. Since the sum of a degree $n$ polynomial is a degree $n+1$ polynomial and the sum of $O(r^k)$ is $c+O(r^k)$, $a_k=P_{n+1}(k)+O(r^k)$.

Iterating, we get that if $\Delta^n a_k=O(r^k)$, then $a_k=P_{n-1}(k)+O(r^k)$.

We can do this by induction on $m$ where $m$ is the order of the pole.

For a simple pole assume we have proved the result.

Then suppose $m>1$, then $(z-z_0)^mf(z)=g(z)$ where $g(z)$ is holomorphic on a disk of radius bigger than $1$.

Consider $h_i(z):= (z-z_0)^{m-i}f(z)$ for $1\leq i \leq m$.
Each $h_i$ is holomorphic on $B(0,1)$ and has a pole of order $i$ at $1$.

$h_i(z)= \sum_{n=0}^{\infty} a_{in} z^n, |z|<1$

Since $h_m(z)=f(z)$ we would be done if we show $\lim_{n\to \infty} \frac{a_{in}}{a_{i(n+1)}}=z_0 \Rightarrow \lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}}=z_0$
Note: We use the induction hypothesis to get this for $h_1$.

To this end, we make the following observation.


Comparing coefficients we deduce


and hence the following

$a_{(i+1)n}z_0^{n+1}= -(a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n)$

Now, $\lim_{n\to \infty} \frac{a_{(i+1)n}}{a_{(i+1)n}} = z_0 \lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}$.

Thus it is enough to show that

$\lim_{n\to \infty} \frac{a_{(i+1)n}z_0^n}{a_{(i+1)n}z_0^{n+1}}= 1$

$\lim_{n\to \infty} \frac{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n}{a_{i0}+a_{i1}z_0 + a_{i2} z_0^2 + \ldots + a_{in}z_0^n + a_{i(n+1)}z_0^{n+1}}=1$

$\lim_{n \to \infty} \frac{S_n(z_0)}{S_{n+1}(z_0)}=1$

where $S_n(z)$ is the $n^{th}$ partial sum at $z$ of $h_i$.

But the above limit is same as
$\lim_{n \to \infty} \lim_{z\to z_0} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)} $

Then since the inner term converges uniformly in $z$ (independent of $n$) we can interchange the limits, which gives

$ \lim_{z\to z_0} \lim_{n \to \infty} \frac{(z-z_0)^iS_n(z)}{(z-z_0)^iS_{n+1}(z)}=\lim_{z \to z_0} \frac{g(z)}{g(z)}=1$

Please let me know if anything is unclear.