I’m trying to solve the following problem, but I can’t. I need your help.
Recall (Sec. 11) that a point $z$ is an accumulation point of a set $S$ if each deleted neighborhood of $z$ contains at least one point of $S$. One form of the Bolzano–Weierstrass theorem can be stated as follows: an infinite set of points lying in a closed bounded region $R$ has at least one accumulation point in $R$. Use that theorem and Theorem 2 in Sec. 75 to show that if a function $f$ is analytic in the region $R$ consisting of all points inside and on a simple closed contour $C$, except possibly for poles inside $C$, and if all the zeros of $f$ in $R$ are interior to $C$ and are of finite order, then those zeros must be finite in number.
My attempt is as follows
Suppose there are infinitely many zeros.
Then by Bolzano-Weierstrass theorem, there is a point $z\in R$ such that every deleted neighborhood of $z$ contains at least one zero.
Then there are two cases.
- $f$ is analytic at $z$
- $z$ is a pole
I solved Case 1. Since $f$ is continuous at $z$, $f(z)=0$.
Then by theorem 2 in Sec. 75, which states that if an analytic function $f$ is not zero function near a zero then the zero is isolated, $z$ has a deleted neighborhood that does not contain any zeros. This contradicts that $z$ is an accumulation point of zeros.
How can I deal with Case 2?
Best Answer
If $f$ has infinitely many zeros in a compact region $K$ then, since $K$ is compact, there is a sequence $(z_n)_{n\in\mathbb N}$ of those zeros which converges to some $z_0\in K$. By the continuity of $f$, $f(z_0)=0$ then. But it follows from this that the set of zeros of $f$ has an accumulation point (which is $z_0$) and therefore, by the identity theorem, $f$ would be the null function.