4

enter image description here

I am reviewing for an Analysis qual and stumbled upon this question. In particular, I am having difficulties with part (ii). My attempt is the following:

Using the hint, let $\Omega = \mathbb{C}$, $S=\{1/n : n\in \mathbb{N}\}$, and $g(z)=z^2$. We have that since $S \subset \Omega$ and both $f$ and $g$ are entire, then $f$ and $g$ are analytic on $S$. Per the uniqueness result, if $g(z)=f(z)$ for all $z\in S$, we know that since $0$ is a limit point of $S$ that is in $\Omega$, then it must be the case that $g(z)=f(z)$ for all $z\in \Omega$. However, we are given that $|f(i)| =2$, yet $|g(i)| = 1$. So in this case, just because $|g(z)| = |f(z)|$ for all $z\in S$, we don't have $g(z)=f(z)$. My strategy is then to find different functions $g$ such that $|g(z)| = |f(z)|$ for all $z\in S$ and $|g(i)|=2$. After finding all these different $g's$, I should have all the possible values of $|f(-i)|$ by just calculating $|g(-i)|$. However, I'm having trouble finding even a single function $g$ that satisfies these two conditions, much less finding all of them. Is there some systematic way I can go about finding these different $g$ functions?

French Toast Crunch
  • 575

1 Answers1

4

First, if $f$ is any entire function, $\overline{f(\bar{z})}$ is always entire, because $\overline{\sum_{n=0}^\infty a_n \bar{z}^n} = \sum_{n=0}^\infty \bar{a_n}z^n,$ which converges exactly when $\sum_{n=0}^\infty a_n z^n$ does.

As $f$ is holomorphic at $0$ and not uniformly $0$, there is a unique integer $n \geq 0$ so $\lim_{z \to 0} \frac{f(z)}{z^n}$ is a nonzero complex number ($n$ is the order of the zero of $f$ at $0$). Our hypotheses show that $f$ has a zero of degree $2$ at $0$. So we can express $f(z) = z^2 h(z)$ for some entire function $h$, and our hypotheses show that $|h(z)| = 1$ for $z = 1/n, n \geq 1$. In particular, $|h(0)| = 1$. So there's a neighborhood of $0$ on which $h(z)$ is nonzero.

Now put $g(z) = \frac{1}{\overline{h(\bar{z})}}$, which holomorphic on a neighborhood of $0$, by part i. We observe that for $1/n$, $n\geq 1$, $$\frac{h(z)}{g(z)} = h(z) \overline{h(z)} = |h(z)|^2 = 1.$$ So by the identity principle, $g(z)$ and $h$ agree in a neighborhood of $0$, and we discover that in fact on all $z \in \mathbb{C}$, $h(z) = \frac{1}{\overline{h(\bar{z})}},$ or equivalently that $h(\bar{z}) = \frac{1}{\overline{h(z)}}$. For $|z| = 1$, $|f(z)| = |h(z)|$ and we see that for $|z| = 1$, $|f(\bar{z})| = \frac{1}{|f(z)|}$. So we conclude $|f(-i)| = 1/|f(i)| = 1/2.$

Alex Nolte
  • 4,986
  • 1
  • 12
  • 26
  • I follow your argument. To clarify, you get that f has a zero of degree $2$ from the fact that we're given $|f(1/n)| =1/n^2$ and then we can take a limit as $n$ tends to infinity, correct? I ask because that first statement of the second paragraph: "As $f$ is holomorphic as $0$ and not uniformly at $0$" threw me off a bit. I don't believe I know what uniformly holomorphic is, but it doesn't seem like we really used that part, unless I missed where it was used. – French Toast Crunch Jun 17 '20 at 01:20
  • 1
    That is how we checked the degree of the zero - we divide $f$ by $z^n$ and the first $n$ where there's a sequence $z_k \to 0$ where $\lim_{k \to \infty} f(z_k)/(z_k^n)$ is a nonzero complex number is the degree of the zero. Mentioning that $f$ isn't uniformly zero is ruling out that $f(z) = 0$ for all $z \in \mathbb{C}$, in which case it doesn't make sense to talk about the order of the zeros of $f$ (this is a hypothesis needed to factor out the $z^2$). – Alex Nolte Jun 17 '20 at 01:38