# An analytic function in a compact region has finitely many zeros

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.

1. $$f$$ is analytic at $$z$$
2. $$z$$ 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?

• If $z$ is a pole, for some integer $M > 0$, $g(w)=f(w)(w-z)^M$ is analytic near $z$, and has zeroes accumulating around $z$. – Mindlack Sep 11 '19 at 8:19
• Umm... Sorry I don’t understand. Could you explain more? The function g is not zero at z ? – anadad Sep 11 '19 at 8:34
• It is, if $M$ is large enough. – Mindlack Sep 11 '19 at 8:38
• So g is analytic and nonzero at z. But z is an accumulation point of zeros. So g(z) must be zero. So this contradicts g is nonzero at z? – anadad Sep 11 '19 at 8:47

If $$f$$ has a pole at $$z$$ then $$|f (\zeta)| \to \infty$$ as $$\zeta \to z$$. Hence there is a deleted neighborhood in which $$f$$ has no zeros.

• Thanks! But could you explain more why there is such deleted neighborhood? – anadad Sep 11 '19 at 8:26
• There exists $\delta >0$ such that $|f(\zeta)| >1$ whenever $0<|\zeta -z| <\delta$. So $f(\zeta) \neq 0$ for such $\zeta$. – Kavi Rama Murthy Sep 11 '19 at 8:29
• Oh! Right! Thanks! But I’m little bit worried. I learned f goes infinity as ζ goes z in section 77. But this exercise is from section 76. – anadad Sep 11 '19 at 8:37

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.

• Thanks! But I don’t understand why f is continuous at z0. If z0 is a pole, f is not analytic. – anadad Sep 11 '19 at 8:25
• I am assuming that $K$ is a subset of the domain of $f$. Besides, a sequence of zeros cannot converge to a pole because, if $f$ has a pole at $z_0$, then $\lim_{z\to z_0}\bigl\lvert f(z)\bigr\rvert=\infty$. – José Carlos Santos Sep 11 '19 at 8:27
• Umm I think I still don’t get why a sequnce of zeros cannot converge to a pole. I wanted to show this when I’m trying to this exercise, but I couldn’t. Could you explain it more? – anadad Sep 11 '19 at 8:44
• If $z_0$ is a pole of $f$, then $\lim_{z\to z_0}\bigl\lvert f(z)\bigr\rvert=\infty$. Therefore, there is a $\varepsilon>0$ such that $\lvert z-z_0\rvert<\varepsilon\implies\bigl\lvert f(z)\bigr\rvert>1$. In particular, $\lvert z-z_0\rvert<\varepsilon\implies f(z_0)\neq0$. In other words, $f$ has no zeros in the open disk centered at $z_0$ with radius $\varepsilon$. – José Carlos Santos Sep 11 '19 at 8:47
• Do you assume f is continuous at z0? – anadad Sep 11 '19 at 8:51