The following is the question:
The function $1/\sin z$ is expressed as a power series around the point $z_0 = 4i + \pi/2$. Explain why the radius of convergence cannot exceed $R = \dfrac{1}{2}\sqrt{64 + \pi^2}$.
My attempt:
The distance of $0$ from $z_0$ is $R$. Thus, if the radius of convergence exceeded $R$, then $0$ would be within the disc of convergence. The function $1/\sin z$ blows up as $z\to 0$ and thus, this would pose a problem.
However, there is the following possibility that I would like to exclude:
The power series agrees with $1/\sin z$ on some disc of $r < R$ around $z_0$ but still converges on a disc of radius $R' > R$ without agreeing with $1/\sin z$ outside the disc of radius $r$.
Now, by using something such as the Identity theorem, it is clear that the above possibility cannot happen. However, I came across notes where this exercise was given at a more elementary level. (Just after the introduction of power series.) Thus, I'm looking for a more elementary solution.
EDIT: To further clarify what I meant-
I wish to show that the following is not possible:
There exists a sequence $(a_n)_{n=0}^\infty$ of complex numbers and $r > 0$ such that:
$$\sum_{n=0}^\infty a_n(z – z_0)^n = \dfrac{1}{\sin z}$$
for all $|z – z_0| < r$.
Furthermore, the radius of convergence of the power series $\sum a_n(z – z_0)^n$ is greater than $R$.
Best Answer
Adding an answer myself. The idea is to use the following fact:
We proceed as follows:
Define $f(z) := 1/\sin z$ for $z \in \Bbb C\setminus\{n\pi \mid n \in \Bbb Z\}$.
Let $D = \{z \in \Bbb C \mid |z - z_0| < R\}$. ($z_0$ and $R$ being as in the question.)
It is clear that $D$ is the maximal disc as mentioned in the fact.
Suppose that the power series converged on a strictly larger disc $D'$.
Let $P:D'\to\Bbb C$ denote the function defined by the power series. Also, note that $P$ is continuous.
We have $P|_D = f|_D$. (By the fact.)
Then, $0 \in {\rm int}\ D'$ and thus, the power series converges (absolutely) at $0$.
Let $(z_n)$ be a sequence in $D$ such that $z_n \to 0$. (Such a sequence exists as $0 \in \bar{D}$.)
Since $P$ is continuous, we must have $$P(z_n) \to P(0).$$ On the other hand, we have $P(z_n) = f(z_n)$ and thus, $$P(z_n) = f(z_n) = \dfrac{1}{\sin z_n},$$ which diverges since $z_n \to 0$.
The above is a contradiction.