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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?

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  • $\begingroup$ 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 $z$. $\endgroup$ – Mindlack Sep 11 '19 at 8:19
  • $\begingroup$ Umm... Sorry I don’t understand. Could you explain more? The function g is not zero at z ? $\endgroup$ – anadad Sep 11 '19 at 8:34
  • $\begingroup$ It is, if $M$ is large enough. $\endgroup$ – Mindlack Sep 11 '19 at 8:38
  • $\begingroup$ 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? $\endgroup$ – anadad Sep 11 '19 at 8:47
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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.

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  • $\begingroup$ Thanks! But could you explain more why there is such deleted neighborhood? $\endgroup$ – anadad Sep 11 '19 at 8:26
  • $\begingroup$ There exists $\delta >0$ such that $|f(\zeta)| >1$ whenever $0<|\zeta -z| <\delta$. So $f(\zeta) \neq 0$ for such $\zeta$. $\endgroup$ – Kavi Rama Murthy Sep 11 '19 at 8:29
  • $\begingroup$ 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. $\endgroup$ – anadad Sep 11 '19 at 8:37
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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.

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  • $\begingroup$ Thanks! But I don’t understand why f is continuous at z0. If z0 is a pole, f is not analytic. $\endgroup$ – anadad Sep 11 '19 at 8:25
  • $\begingroup$ 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 $\lim_{z\to z_0}\bigl\lvert f(z)\bigr\rvert=\infty$. $\endgroup$ – José Carlos Santos Sep 11 '19 at 8:27
  • $\begingroup$ 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? $\endgroup$ – anadad Sep 11 '19 at 8:44
  • $\begingroup$ 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 $\varepsilon$. $\endgroup$ – José Carlos Santos Sep 11 '19 at 8:47
  • $\begingroup$ Do you assume f is continuous at z0? $\endgroup$ – anadad Sep 11 '19 at 8:51

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