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While studying for my complex analysis course, I came across the following question:

Let $\Omega$ be an open subset of $\mathbb{C}$ and let $a \in \Omega$. Assume that $f : \Omega \setminus \{a\} \to \mathbb{C}$ is holomorphic with $$ \lim_{z \to a} (z - a) f(z)^2 = 0 $$ Prove that $f$ has a removable singularity at $a$.

I am familiar with the Riemann's theorem on removable singularities, so I was thinking about proving the fact that the function $f^2$ has a removable singularity at $a$ with this theorem and then taking the square root of this function.

However, I am not sure if the square root of a holomorphic function with a removable singularity at $a$ is still a holomorphic function with a removable singularity at $a$. I can't seem to find another way of proving this question. Can anyone help me with this please?

Lucius Aelius Seianus
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3 Answers3

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You have\begin{align}\lim_{z\to a}(z-a)f^2(z)=0&\implies\lim_{z\to a}(z-a)^2f^2(z)=0\\&\iff\lim_{z\to a}\left((z-a)f(z)\right)^2=0\\&\iff\lim_{z\to a}(z-a)f(z)=0.\end{align}So, by Riemann's extension theorem, you can extend $(z-a)f(z)$ to a holomorphic function $F\colon\Omega\longrightarrow\Bbb C$. Besides,$$F(a)=\lim_{z\to a}(z-a)f(z)=0.$$So, the Taylor series of $F$ centered at $a$ is of the type $a_1(z-a)+a_2(z-a)^2+\cdots$, and therefore, near $a$,$$f(z)=a_1+a_2(z-a)+a_3(z-a)^2+\cdots$$

José Carlos Santos
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Rather than directly applying Riemann’s theorem (because square roots are annoying), I would rather mimic its proof via Cauchy’s inequalities. We know $f$ has a Laurent expansion in some punctured disk around $a$, say $f(z)=\sum\limits_{n=-\infty}^{\infty}c_n(z-a)^n$. We know the coefficients are \begin{align} c_n=\frac{1}{2\pi i}\int_{|z-a|=\epsilon}\frac{f(z)}{(z-a)^{n+1}}\,. \end{align} Our hypothesis on $f$ implies that $(z-a)^2f(z)$ is bounded near $a$, and hence there is a $B>0$ such that for all small enough $\epsilon$, if $0<|z-a|\leq \epsilon$, then $|f(z)|\leq \frac{B}{\sqrt{|z-a|}}$. Thus, \begin{align} |c_n|\leq \frac{1}{2\pi}\int_{|z-a|=\epsilon}\frac{B}{\sqrt{|z-a|}}\frac{1}{|z-a|^{n+1}}|dz|=\frac{1}{2\pi}\cdot\frac{B}{\sqrt{\epsilon}}\cdot\frac{1}{\epsilon^{n+1}}\cdot 2\pi\epsilon=\frac{B}{\epsilon^{n+\frac{1}{2}}}. \end{align} Hence, if $n\leq -1$, then $|c_n|\leq B\epsilon^{|n|-\frac{1}{2}}\to 0$ as $\epsilon\to 0^+$, proving that $c_n=0$. Thus, we have $f(z)=\sum_{n=0}^{\infty}c_n(z-a)^n$, proving that $f$ extends to be holomorphic at $a$.

This sort of technique of directly estimating the integral is very flexible, and allows you to prove holomorphic extension theorems/Louiville-type theorems very easily. This is one of the powers of integrals. See here and here for more illustrations of this idea. The general idea is that growth/decay assumptions on $f$ imply corresponding properties about various terms in the Laurent expansion of $f$.

peek-a-boo
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Let $g(z)=(z-a)f(z), z \neq a, g(a)=0$. Note that $|g(z)| \to 0$ as $z \to a$ by hypothesis. [$|f(z)|<\frac 1 {\sqrt {|z-a|}}$ for $|z-a|$ sufficiently small].

Thus, $g$ is analytic at $a$ and $g$ has a zero at $a$. So $g(z)=(z-a)^{k}h(z)$ with $h$ analytic at $a$, $k$ being the order of zero of $g$ at $a$. Can you finish?

Kavi Rama Murthy
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