Suppose $f(z)$ is an analytic function in a region $D$, and $f(z)$ is not identically equal to zero there. Let $z_n$ be a sequence of zeros of $f(z)$ in $D$. If $\lim_{n\to \infty}z_n=a$, then $a$ must be singularity of $f(z)$ (As zeros of analytic functions are isolated). I would like to classify the type of singularities that $f(z)$ can have at point $a.$
I could show that $a$ is a singularity. But I was not sure the type of the singularity. I had a part which was wrong (Saying that the singularities of Analytic function are isolated) and I took it off. Sorry if I am asking too many questions, I am new in complex analysis but I think this question was a natural and good! Thanks.
Best Answer
It can't be a removable singularity. If it was, $\lim_{n\to\infty}f(z_n)=0$ and therefore we could extend $f$ to an analytical function whose domain contains $a$ and, in that extension, $a$ would be a non-isolated zero. Therefore, $f$ would be the null function.
And it cannot be a pole either, because then we would have$$\lim_{n\to\infty}z_n=a\implies\lim_{n\to\infty}\bigl|f(z_n)\bigr|=+\infty,$$which is not true since, in fact, $\lim_{n\to\infty}f(z_n)=0$.
So, it can only be an essential singularity, such as when $f(z)=\sin\left(\frac1z\right)$.